WBBSE Solutions For Class 10 Maths Algebra Chapter 1 Quadratic Equation In One Variable

Algebra Chapter 1 Quadratic Equation In One Variable What Is Meant By A Quadratic Equation In One Variable

WBBSE Class 10 Quadratic Equations Overview

In class 9, you have studied a lot about polynomials in one variable. If in a polynomial, there is only one variable (i.e., x or y or z or any other variable), then it is called a polynomial in one variable.

For example, 4x + 7, 3y + 10, 10z + 1, etc. Now if the highest power of this variable (i.e., power of x, y, or z) be 2, then it is called a quadratic polynomial in one variable.

Such as x² + 2x + 1, 3y² + 7y + 10, 2z² + 6z + 8, etc.

When equations are formed by these quadratic polynomials, then that equations are called quadratic equations in one variable.

Such as, x² + 3x + 2 = 0, 4y² + 5y – 6 = 0, etc.

WBBSE Solutions for Class 10 Maths

The standard form or General form of a quadratic equation in one variable

ax² + bx + c – 0, where a, b, c are reals and a ≠0.

So, the equations which can be expressed in the form ax² + bx + c = 0, where a, b, c are reals and a≠0 are called quadratic equations in one variable with real coefficients.

The characteristics of a quadratic equation in one variable are

There will be only one variable;

The highest power of the variables must be 2;
The coefficients of the variables are all real; and
The coefficient of the term consisting of the quadratic variable must be a non-zero real number.

To solve any real problem, we can express the problem in a quadratic equation in one variable, such as,

The length of a rectangular garden is 36 meters more than its breadth.

If the area of the garden be 460 square meters, then find the length and breadth of the garden.

To solve the above real problem, we can express the problem as a quadratic equation in one variable. For example,

Let the breadth of the garden be x m.

∴ Length = (x+36)m

∴ Area of the garden = Length x Breadth

= (x + 36) x x sq.m. = (x² + 36x) sq.m.

As per question, x² + 36x = 460

or, x² + 36x – 460 = 0, which is a quadratic equation in one variable (say x) with real coefficients.

By solving this equation we can determine the length and breadth of the garden.

Algebra Chapter 1 Quadratic Equation In One Variable What Is Meant By A Quadratic Equation In One Variable Multiple Choice Questions

Example 1. For what value of a, the equation (a – 2) x² + 3x + 5 = 0 will not be a quadratic equation?

Solution: The given equation, (a- 2) x² + 3x +5 = 0 will not be a quadratic equation if a – 2 = 0 (i.e., the coefficient of x² is equal to 0), or, a = 2.

∴ 2. is correct.

Value of a is 2

Example 2. For what power of the variable of the equation x⁶-x³ – 2 = 0 will be a quadratic equation?

Solution: x⁶– x³ – 2 = 0 or, (x³)² – x³ – 2 = 0

or, y² – y – 2 (when x³ = y) = 0

Now, y²-y-2 = 0 is a quadratic equation.

∴ The given equation is a quadratic equation with respect to x³

i.e, for the power 3 of the variable it is a quadratic equation.

∴ 3. is correct.

Example 3. If the expression = \(\frac{x}{4-x}=\frac{1}{3 x}\) (x≠0, x≠4) is expressed in the form of ax² + bx + c = 0 (a≠0), then the co-efficient of x will be.

Solution: Given that \(\frac{x}{4-x}=\frac{1}{3 x}\)

or, 3x² = 4 – x or, 3x² + x – 4 = 0

∴ the co-efficient of at is equal to 1.

∴ 1 is correct.

Maths Solutions Class 10 Wbbse

The co-efficient of x will be 1

Example 4. Which of the following is a quadratic polynomial?

x³– 7x + 2
4x⁵ -x (x + 2)
2x(x + 5) + 1

2x – 1

Solution: In the polynomial x³– 7x + 2, the highest power of the variable x is 3

∴ it is not a quadratic polynomial.

In the polynomial 7x- x (x + 2), the highest power of the variable x is 5.

∴ it is also not a quadratic polynomial.

The polynomial 2x (x + 5) + 1 = 2x² + 10x + 1 can be expressed in the form ax²+ bx + c, where a≠0.

∴ it is a quadratic polynomial.

In the polynomial 2x – 1, the highest power of the variable x is 1. So, it is not a quadratic polynomial.

∴ 2x(x + 5) + 1 is correct.

2x(x + 5) + 1 is a quadratic polynomial

Example 5. Write true or false

1. The equation x + \(\frac{1}{x}\)= 2 (x≠1) is not a quadratic equation in one variable.

Solution: False

since the given equation is x + \(\frac{1}{x}\) = 2

⇒ \( \frac{x^2+1}{x}\)=2

⇒ x²+1=2 x

⇒ x²-2 x+1=0

Which is a quadratic equation in one variable.

Hence the given statement is False.

2. The equation (x -1)³ = x (x² – 1) iis a quadratic equation in one variable.

Solution: True

Since the given equation is (x – 1)³ = x (x² – 1)

⇒ x³ – 3x² + 3x – 1 = x³ – x

⇒ – 3x² + 3x + x – 1 = 0

⇒ – 3x² + 4x – 1 = 0

⇒ 3x² – 4x + 1 = 0

which is the quadratic equation in one variable

Hence the given equation is True.

Example 6. Fill in the blanks

1. The co-efficient of x³in the equation (x + 1)³ = x (x² + 1) is ______

Solution: 0

Since (x+ 1)³= x (x²+ 1)

⇒ x³² + 3X² + 3x + 1 = x³ + x

⇒ 3x² + 2x + 1 =0, where x³ does not belong

2. The given equation x^-1 +x = t(t = constant) is a _______ equation in _______ variabler

Solution: quadratic, one

since x^-1+ x = t

⇒ \(\frac{1}{x}\) + x = t

⇒ x²-tx+1=0

3. The quadratic equation ax²+ bx + c = 0 becomes a linear equation when a = _______

Solution: 0

since if a =0, then a² + bx + c = 0 becomes bx + c, which is a linear

Algebra Chapter 1 Quadratic Equation In One Variable What Is Meant By A Quadratic Equation In One Variable Short Answer Type Questions

Understanding Quadratic Equations in One Variable

Example 1. Express the term 3x² + 7x+ 23 = (x + 4)(x + 3) + 2 in the form of ax + bx +c = 0 (a≠0).

Solution: Given that Sx²+ 7x + 23 = (x + 4)(x + 3) + 2 *

or, 3x² + 7x + 23 = x² + 4x + 3x + 12 + 2

or, 3X² + 7x + 23 = x² + 7x + 14

or, 2x² + 9 = 0 or, 2x² + 0.x + 9 = 0

∴ the required quadratic equation of the form ax² + bx + c = 0 (a ≠0) is 2x² + 0.x + 9 = 0.

Example 2. Express the term (x + 2)³= x (x² – 1) in the form ax² + bx + c – 0 (a ≠ 0) of a quadratic equation and find the coefficient of x², x, and x⁰.

Solution: Given that (x + 2)³= x (x² – 1)

or, (x)³ + 3.x².2 + 3. x.2² + 2³ = x³ – x .

or, x³ + 6x² + 12x + 8 = x³ – x

or, x³ + 6x² + 12x + 8 – x³ + x = 0

or, 6X² + 13x + 8 = 0

∴ the required quadratic equation is 6X² + 13x + 8 = 0.

Here, coefficient of x² = 6, coefficient of x = 13 and coefficient of x° = 8.

Example 3. Form a quadratic equation in one variable from the following statement(s) : The product of two consecutive positive odd numbers is 143.

Solution: Let n be any natural number.

Then (2n + 1) is an odd positive number. We know that the difference between two consecutive odd positive numbers is always 2.

∴ the next or the previous odd positive number of (2n+ 1) is (2n + 1 – 2) = 2n – 1

or, (2n + 1 + 2) = 2n + 3.

As per question, (2n – 1)(2n + 1) = 143 or, (2n + 1)(2n+ 3) = 143

or, (2n)² – (1)² = 143 or, 4n² + 2n + 6n + 3 = 143.

or, 4n² – 1= 143 or, 4n² + 8n = 143 -3

or, 4n² – 144 = 0 or, 4n² + 8n – 140 = 0

or, n² – 36 = 0 or, n²+ 2n – 35 = 0

Hence, the required quadratic equations are n² – 36 = 0 and n² +2n-35-0.

Example 4. Express the term x-1+\(\frac{1}{6}\) = 6(x≠0) in the form ax² + bx + c where a, b, c are real numbers and a ≠0.

Solution: \(x-1+\frac{1}{x}\)=6

or, \(\frac{x^2-x+1}{x}\)=6

or, x²– x + 1 = 6x

or, x²– x + 1 – 6x = 0

or, x²– 7x + 1 = 0

Hence the required form is x²– 7x + 1 = 0

Algebra Chapter 1 Quadratic Equation In One Variable What Is Meant By A Quadratic Equation In One Variable Long Answer Type Questions

Example 1. Verify whether the following equation can be written in the standard form or not \(x+\frac{3}{x}=x^2\) (x≠0)

Solution: \(x+\frac{3}{x}=x^2\)

or, \(\frac{x^2+3}{x}=x^2\)

or x² + 3 = x³

or x² + 3 – x³ = 0

or -x³– x² – x = 0

∴ the given equation can not be expressed as the standard form.

Example 2. The sum of the squares of two consecutive numbers is 313.

Solution:

Given:

The sum of the squares of two consecutive numbers is 313

Let x be any number.

∴ The next or the previous number of .y is (a + 1) and (x – 1) respectively.

As per the question, (x – 1)² + x² = 313 and x² + (x + 1)² = 313.

or, x² – 2x +1 + x² = 313 or, x² + x² + 2.y +1 = 313

or, 2x² -2x-312 = 0 or, 2x² + 2x -312 = 0.

or, x²– x – 156 = 0

or, x²+ x- 156 = 0.

Hence, the required quadratic equations in one variable are x²-x- 156 = 0 or, x²+ x – 156 = 0.

Standard Form of Quadratic Equations

Example 3. The diagonal of a rectangular field is 15 m and its length is 3 m more than its breadth.

Solution:

Given:

The diagonal of a rectangular field is 15 m and its length is 3 m more than its breadth

Let the breadth of the field be x m.

∴ length = (x + 3) m

∴ length of the diagonal = \(\sqrt{x^2+(x+3)^2}\) m (as per formula)

As per question, \(\sqrt{x^2+(x+3)^2}\) = 15

or, x² + (x+ 3)² = (15)² (Squaring both sides)

or, x² + x + 6x + 9 = 225

or, 2x² + 6x + 9 – 225 = 0

or, 2x²+ 6x – 216 = 0

or, x² + 3x – 108 = 0

Hence the required quadratic equation in one variable is x² + 3x – 108 = 0.

Example 4. The distance between two stations is 300 km. A train went to the second station from the first station with uniform velocity. If the velocity of the train had been 5 km/hour more, then the time taken by the train to reach the second station would be lesser by 2 hours.

Solution:

Given:

The distance between two stations is 300 km. A train went to the second station from the first station with uniform velocity. If the velocity of the train had been 5 km/hour more

Let the uniform velocity of the train be a km/hour.

∴ the time taken by the train to reach the second station is \(\frac{300}{x}\) hours.

It the uniform velocity of the train had been 5 km/hour more, the time taken by the train to reach the second station is \(\frac{300}{x+5}\) hours.

As per question,

\(\frac{300}{x}-\frac{300}{x+5}=2 \)

or, \(\frac{150}{x}-\frac{150}{x+5}=1\)

or, \(\quad \frac{150 x+750-150 x}{x(x+5)}=1\)

or, \(\frac{750}{x^2+5 x}=1\)

or, x²+5 x= 750

or, x²+5 x-750

Hence the required quadratic equation is x²+5 x-750 = 0

Example 5. One person bought some kg of sugar ₹80. If he would get 4 kg sugar more with that money, then the price of 1 kg sugar would be less by ₹1

Solution:

Given

One person bought some kg of sugar ₹80. If he would get 4 kg sugar more with that money

Let the person bought x kg of sugar in ₹80

∴ the rate of value of sugar = ₹\(\frac{80}{x}\)/kg.

If he would get 4 kg sugar more, then the rate of value of sugar would be ₹\(\frac{80}{x+4}\)/ kg.

As per question,

\(\frac{80}{x}-\frac{80}{x+4}=1\)

or, \(\frac{80 x+320-80 x}{x(x+4)}=1\)

or, \(\frac{320}{x^2+4 x}=1\)

or, x² + 4x – 320 = 0.

Hence the required quadratic equation is x² + 4x – 320 = 0.

Methods to Solve Quadratic Equations

Example 6. A clock-seller sold a clock by purchasing it at ₹336. The amount of his profit percentage is as much as the amount with which he bought the clock.

Solution:

Given

A clock-seller sold a clock by purchasing it at ₹336. The amount of his profit percentage is as much as the amount with which he bought the clock.

Let the cost price of the clock = ₹x.

∴ Profit = ₹(336 – x) [selling price = ₹336]

∴ Percentage of profit = \(\frac{336-x}{x} \times 100 \%\)

As per the question,

\(\frac{336-x}{x} \times 100 \%=x \%\)

or, \(\frac{33600-100 x}{x}=x \)

or, x²= 33600 – 100x

or, x² + 100x – 33600 = 0

Hence the required quadratic equation is x² + 100x – 33600 = 0.

Example 7. If the velocity of the stream is 2 km/hr, then the time taken by Ratanmajhi to cover 21 kms in downstream and upstream is 10 hours.

Solution:

Given

If the velocity of the stream is 2 km/hr,

Let the velocity of the boat of Ratanmajhi = x km/hr.

∴ Velocity of the boat in downstream = (x + 2) km/hr and in upstream = (x – 2) km/hr.

∴ to reach 21 km in downstream, the time required = \(\frac{21}{x+2}\) hours

and to come back, the time required = \(\frac{21}{x-2}\) hours

As per question,

\(\frac{21}{x+2}+\frac{21}{x-2}=10\)

or, \(21\left(\frac{1}{x+2}+\frac{1}{x-2}\right)=10\)

or, \(21\left\{\frac{x-2+x+2}{(x+2)(x-2)}\right\}=10\)

or, \(21\left\{\frac{2 x}{x^2-2^2}\right\}=10\)

or, \(\frac{42 x}{x^2-4}=10\)

or, \( \frac{21 x}{x^2-4}=5 \)

or, 5 x² -21x – 20 = 0

Hence the required quadratic equation is 5x² -21x – 20 = 0

Example 8. The time taken to clean out a garden by Majid is 3 hours more than that by Mahim. Both of them together can complete the work in 2 hours.

Solution: Let the time taken by Mahim to clean out the garden be x hours.

∴ Time taken by Majid = (x + 3) hrs.

Now, in 2 hrs. Mahim works \(\frac{2}{x}\) part of the work and in 2 hrs.

Majid works \(\frac{2}{x+3}\) part of the work.

As per question, \(\frac{2}{x}+\frac{2}{x+3}=1\) {complete work = 1]

or, \(2\left(\frac{1}{x}+\frac{1}{x+3}\right)=1\)

or, \(2\left\{\frac{x+3+x}{x(x+3)}\right\}=1\)

or, \(2\left\{\frac{2 x+3}{x^2+3 x}\right\}=1\)

or, \(\frac{4 x+6}{x^2+3 x}=1\)

or, x²+ 3x = 4x + 6

or, x²-x – 6 = 0

Hence the required quadratic equation is x²-x – 6 = 0

Example 9. The unit digit of a two-digit number exceeds its tens digit by 6 and the product of two digits is less by 12 from the number.

Solution: Let the tens digit of the number = x

∴ unit digit = (x + 6)

The number = x x 10 + x + 6 = 11x + 6.

As per question, x (x + 6) = 11x + 6 – 12

or, x² + 6x = 11x – 6.

or, x² + 6x-11x +6 = 0

or, x² – 5x + 6 = 0.

Hence the required quadratic equation is x² – 5x + 6 = 0.

Quadratic Formula Derivation

Example 10. There is a road of equal width around the outside of a rectangular playground having a length 45 m and breadth 40 m and the area of the road is 450 sq-m.

Solution: Let the breadth of the road = x m.

∴ the length of the field including road = (45 + x + x) m = (2x + 45) m.

and the breadth of the field including road = (40 + x + x) m = (2x + 40) m.

∴ Area of field including road = (2x + 45)(2x + 40) sq-m.

and area of field excluding road = 45 x 40 sq-m = 1800 sq-m.

∴ area of the road, = {(2x + 45)(2x + 40)- 1800} sq-m. . .

As per question, {(2x + 45)(2x + 40) – 1800} = 450.

or, 4x² + 90x + 80x + 1800- 1800 = 450

or, 4x² + 170.x-450 = 0

or, 2x² + 85x – 225 = 0

Hence the required quadratic equation is 2x² + 85x – 225 = 0.

Algebra Chapter 1 Quadratic Equation In One Variable Root Of A Quadratic Equation

We know that the general form of a quadratic equation in one variable is ax² + bx + c = 0, where a, b, c are real numbers and a≠0.

By the term ‘roots of this equation,’ we generally denote those real numbers which satisfy this equation, i.e., if α be a real number such that aα² +bα + c = 0, i.e., α satisfies the equation ax² + bx + c – 0, then α is called a root of the equation ax² + bx + c = 0.

Thus if β be a root of this equation ax² + bx + c = 0, then αβ² + bβ + c = 0.

So, the real numbers or quantities which satisfy a given quadratic equation are called the roots of that equation.

For example, let x² + 3x + 2 = 0 be a given quadratic equation.

Putting x = – 1 in this equation we get, (-1)² + 3 x (-1) + 2 = 1 -3 + 2 = 0; i.e., the real number (- 1), satisfy the quadratic equation x² + 3x+ 2 = 0. .

∴ (- 1) is a root of the equation.

Algebra Chapter 1 Quadratic Equation In One Variable Number Of Roots

In the previous class we have studied that the number of zeroes of a quadratic equation is 2.

So, the number Of roots of the quadratic equations is always 2.

Although the roots may be equal or imaginary. Later on we shall discuss about this.

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of Roots

To determine the roots of a quadratic equation given in standard form, i.e., in the form ax² + bx + c = 0 (where a, b, c are real numbers and a ≠ 0), we generally factorize the quadratic equation into two linear factors and then equalize each of these factors with zero.

From there we determine the value of the variable by simplification. That value of the variable is called the root of the equation.

Again, to verify whether a given number or quantity is a root of the given quadratic equation or not, we examine the equality of LHS and RHS of the equation by putting this number or quantity in stead of the variable in both sides of the equation.

If the LHS and the RHS are equal, then we say that the given number or quantity is a root of the quadratic equation.

For example, let you have to verify whether (- 2) is a root of the quadratic equation 8x² + 7x = 0 or not, then LHS = 8X² + lx = 8 x (- 2)² + 7 x (- 2)

= 32 – 14 = 18

= 18 ≠ 0 = RHS

Thus, LHS and RHS are not equal.

∴ (- 2) is not a root of the given equation 8x + 7x = 0.

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of The Solution Of A Given Quadratic Equation

Determination of the solution of a given quadratic equation means that we want to determine those values of the variable by which the given equation is satisfied.

Since every root of a quadratic equation satisfies the equation, so each root is a solution of the given quadratic equation. For example,

The roots of the quadratic equation 8x² + 7x = 0 is

We have, 8x² + 7x = 0 or, x (8x + 7) = 0

⇒ either x = 0 or, 8x + 7 = 0 .

⇒ 8x = – 7

⇒ x = –\(\frac{7}{8}\)

∴ x = 0 or, x = –\(\frac{7}{8}\)

i.e.., the roots are 0 and –\(\frac{7}{8}\)

So, the solution of the quadratic equation 8x² + 7x – 0 are x = 0 and x = –\(\frac{7}{8}\) and the roots are o and –\(\frac{7}{8}\).

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of The Solution Of A Given Quadratic Equation Multiple Choice Questions

Example 1. The value of k for which \(\frac{2}{3}\) is a root of the quadratic equation 7x² + kx- 3 = 0 is

–\(\frac{1}{6}\)
\(\frac{1}{6}\)
\(\frac{5}{6}\)
–\(\frac{5}{6}\)

Solution: 7x²+kx-3 = 0

or, \(7 \times\left(\frac{2}{3}\right)^2+k \times \frac{2}{3}-3=0\)

or, \(7 \times \frac{4}{9}+\frac{2 k}{3}-3=0\)

or, \(\frac{28+6 k-27}{9}=0\)

or, \(\frac{6 k+1}{9}=0\)

or, 6 k+1=0

or, 6 k=-1

or, \(k=-\frac{1}{6}\)

∴ 1. –\(\frac{1}{6}\)

Example 2. The value of k for which (-a) will be a root of the quadratic equation x²+3ax+k = 0

a²
-a²
-2a²
2a²

Solution: We have, x² + 3ax + k = 0

or, (-a)² + 3 a.(-a) + k- 0 [ – a is a root]

or, a² – 3a² + k = 0

or, – 2a² + k – 0

or, k = 2a²

∴ 1. 2a²is correct.

Examples of Solving Quadratic Equations

Example 3. One root of the quadratic equation x²+ 2x+ 1 =0 is

0
-1
1
None of these

Solution: x²+ 2x + 1 = 0

or, (x + 1)² = 0 or, x + 1 = 0 or,x = – 1

∴ (- 1) is a root of the quadratic equation x² + 2x + 1 =0.

∴ 2. -1 is correct.

One root of the quadratic equation x²+ 2x+ 1 =0 is -1

Example 4. One root of the quadratic equation x²– √3x- 6 = 0 is

4√3
√3
2√3
-2√3

Solution: We have, x²– √3x-6 = 0

or, x² – (2√3 – √3)x-6 = 0

or, x² – 2√3x +√3x -6 = 0 .

or, x(x-2√3) + √3(x-2√3) = 0

or, (x-2√3)(x+√3) = 0

∴ either (x-2√3) = 0 or, (x+√3) = 0

⇒ x = 2√3 ⇒ x = -√3

∴ 2√3 is a root of the given equation.

∴ 3. 2√3 is correct.

One root of the quadratic equation x²– √3x- 6 = 0 is 2√3

Example 5. Write true or false

1. The value of k for which a will be a root of the quadratic equation x² – 3ax – k = 0 is 4a².

Solution: True

Since if a be a root of the given equation x + 3ax – k= 0 then

a² + 3a.a – k = 0 ⇒ a² + 3a² – k = 0 ⇒ 4a² – k = 0 ⇒ k = 4a²

Hence the statement is True.

2. The quadratic equation x²– 2x + 1=0 have no real root.

Solution: False

Since we have x²– 2x + 1 = 0 ‘

or, (x – 1 )² = 0 or, x – 1 = 0 or, x = 1

i.e..,1 is a real root of the given quadratic equation.

Hence the statement is False.

Example 6. Fill in the blanks

1. The solutions of the quadratic equation x + \(\frac{1}{10}\) = \(\frac{10}{3}\) are _____ and ______

Solution: 3 and \(\frac{1}{3}\)

since, 3 + \(\frac{1}{3}\) = \(\frac{10}{3}\) and

\(\frac{1}{3}+\frac{1}{\frac{1}{3}}=\frac{1}{3}+3=\frac{10}{3}\)

Aliter: \(x+\frac{1}{x}=\frac{10}{3}\)

or, \(\frac{x^2+1}{x}=\frac{10}{3}\)

or, 3x² + 3 =10x or, 3x² – 10x + 3 =10

or, 3x² – x – 9x + 3 = 0 or, x (3x – 1) – 3 (3x – 1) = 0

or, (3x – 1) (x – 3) = 0

∴ 3x – 1 = 0 or, x = \(\frac{1}{3}\); and x-3 = 0 or, x = 3

2. If one root of the quadratic equation 3x²-√3x-a = 0 be, √3, then the value of a is _____

Solution: 6

since 3(√5)²-√3.√3 – a= 0

or, 9 – 3 – a = 0 or, 6 – a = 0 or, a= 6

3. If 1 be a root of the equation x + kx^-1 = 2, the value of k is _____

Solution: 1

since x + kx^-1 =2

⇒ 1 + k.1^-1 = 2

or, 1 + k = 2 or, k = 1

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of The Solution Of A Given Quadratic Equation Short Answer Type Questions

Example 1. Solve: x +\(\frac{1}{x}\) = \(\frac{13}{6}\)

Solution: \(x+\frac{1}{x}=\frac{13}{6}\)

or, \(\frac{x^2+1}{x}=\frac{13}{6}\)

or, 6x²+ 6 = 3x

or, 6x²-13x + 6 = 0

or, 6x²-4x -9x + 6 = 0

or, 2x(3x-2) – 3(3x-2) = 0

or, (3x-2)(2x-3) = 0

Either 3x – 2 = 0

⇒3x = 2 ⇒ x = \(\frac{2}{3}\)

2x-3 = 0

⇒2x = 3 ⇒ x = \(\frac{3}{2}\)

∴ The required solution are x = \(\frac{3}{2}\) and x = \(\frac{2}{3}\)

Example 2. If one root of the quadratic equation 3x²+ √2x + a = 0 be √2 then find the value of a.

Solution: √2 is a root of the equation 3x² + √2x + a = ().

∴ 3.(√2)²+ √2. √2+ a= 0 or, 6 + 2 + a = 0 or, a = – 8

∴ The value of a = – 8.

Example 3. Prove that the quadratic equation x²– 6x + 5 = 0 have the roots 1 and 5.

Solution: We have, x²– 6x + 5 = 0

or, x² – (1 + 5)x + 5 = 0

or, x² – x – 5x + 5 = 0

or, x(x-1)-5(x-1)= 0

or, (x – 1)(x – 5) = 0

Either x – 1 = 0 or, x – 5 = 0

⇒ x = 1 ⇒ y = 5.

∴ 1 and 5 are the roots of x² – 6x + 5 = 0 (Proved)

Example 4. For what value of k, 1 is a root of the equation x + \(\frac{k}{x)\) = 2?

Solution: 1 is a root of the equation x + \(\frac{k}{x}\)= 2, we get,

1 + \(\frac{k}{x}\) = 2 or, 1 + k = 2 or, k = 2 – 1 or, k = 1.

∴ The required value of k is 1.

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of The Solution Of A Given Quadratic Equation Long Answer Type Questions

Example 1. If two roots of the quadratic equation ax²+ 7x + b = 0 be \(\frac{2}{3}\) and – 3, then find the value of a and b.

Solution: \(\frac{2}{3}\) is a root of the equation ax²+ 7x + b =0,

We get, \(\left(\frac{2}{3}\right)^2\) + 7 x \(\frac{2}{3}\) + b = 0

or, \(\frac{4a}{9}\) + \(\frac{14}{3}\) + b = 0

or, 4a + 42 + 9b = 0 ….(1)

Again, since (- 3) is a root of the equation ax² + 7x + b = 0,

we get, a x (- 3)² + 7 x (- 3) + b = 0 or, 9a – 21 + b = 0

or, 81a – 189 + 9b = 0 ….. (2)

Now, subtracting (1) from (2) we get, 77a – 231 = 0

or, 77a= 231 or, a=\(\frac{231}{77}\) or, a = 3.

Now putting a = 3 in (1) we get, 4 x 3 + 42 + 9b = 0 or, 12 + 42 + 9b = 0 ,

or, 9b + 54 = 0 or, b =- \(\frac{54}{9}\) or, b = – 6.

∴ The required values of a and b are a = 3, b = – 6.

Example 2. Solve

1. (2x+ 1)²+ (a + 1)² = 6x + 47.

Solution: 1. (2x + 1)² + (x + 1)² = 6x + 47

or, (2x)² + 2.2x. 1 + (1)² + (x)² + 2.x. 1 + (1)² = 6x + 47.

or, 4x² + 4x + 1 + x² + 2x + 1 = 6x + 47

or, 5x² + 6x + 2 = 6x+ 47

or, 5x² = 47-2 or, 5x² = 45 or, x² = 9 or, x = ± 3.

Hence the required solutions are a = 3 and x = – 3.

2. \(3 x-\frac{24}{x}=\frac{x}{3}(x \neq 0)\)

Solution: \(3 x-\frac{24}{x}=\frac{x}{3}(x \neq 0)\)

or, \(\frac{3 x^2-24}{x}=\frac{x}{3}\)

or, 2 – 5x + 2x²= 0

or, 8x² – 72 = 0 or, 8(x² – 9) = 0

or, x² – 9 = 0

or, x² = 9 or, x = ± 3.

Hence the required solutions are x = 3 and x = – 3.

3. \(\frac{2}{x^2}-\frac{5}{x}+2=0(x \neq 0)\)

Solution: \(\frac{2}{x^2}-\frac{5}{x}+2=0(x \neq 0)\)

or, \(\frac{2-5 x+2 x^2}{x^2}=0\)

or, 2- 5x + 2x² = 0

or, 2x² – 5x + 2 = 0

or, 2x²– (1 + 4)x + 2 = 0

or, 2x²– x – 4x + 2 = 0

or, x (2x – 1) – 2 (2x-1) = 0

or, (2x – 1)(x-2) = 0

∴ either 2x- 1 = 0 or, x – 2 = 0

⇒ 2x = 1 ⇒ x = 2

⇒ x = \(\frac{1}{2}\)

Hence the solutions are x = \(\frac{1}{2}\) and x = 2.

4. \(\frac{x-2}{x+2}+6\left(\frac{x-2}{x-6}\right)=1(x \neq-2,6)\)

Solution: \(\frac{x-2}{x+2}+6\left(\frac{x-2}{x-6}\right)=1(x \neq-2,6)\)

or, \(\frac{(x-2)(x-6)+6(x-2)(x+2)}{(x+2)(x-6)}=1\)

or, x² -2x-6x + 12 + 6(x² -2²)= (x + 2)(x-6)

or, x² -8x+ 12 +6x² -24 = x² -2x-6x-12 =

or, 7x² – 8x – 12 – x² – 2x + 6x + 12 =0

or, 6x² – 4x = 0

or, 2x (3x – 2) = 0

∴ either 2x = 0 or, 3x – 2 = 0

⇒ x = 0 ⇒ 3x = 2 ⇒ x = \(\frac{2}{3}\)

Hence the required solution are x = 0 and x = \(\frac{2}{3}\)

5. \(\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{1}{12}(x \neq 0,-1)\)

Solution: \(\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{1}{12}(x \neq 0,-1)\)

or, \(a+\frac{1}{a}=\frac{25}{12}\)

putting \(\frac{x}{x+1}\)=a, \(\frac{x+1}{x}=\frac{1}{a}\)

or, \(\frac{a^2+1}{a}=\frac{25}{12}\)

or, 12a² + 12 = 25a

or, 12a² – 25a + 12 = 0

or, 12a² – (9 + 16) a + 12 = 0 or, 12a² – 9a – 16a + 12 = 0

or, 3a(4a – 3) – 4 (4a – 3) = 0 or, (4a – 3)(3a – 4) = 0

either 4a -3 = 0

⇒ 4a = 3

⇒ a = \(\frac{3}{4}\)

putting a = \(\frac{x}{x+1}\)

⇒ \(\frac{x}{x+1}=\frac{3}{4}\)

⇒ 4x = 3x + 3

⇒ 4x – 3x = 3

⇒ x = 3

or, 3a-4 = 0

⇒ 3a = 4

⇒ a = \(\frac{4}{3}\)

[putting a = \(\frac{x}{x+1}\)]

⇒ \(\frac{x}{x+1}=\frac{4}{3}\)

⇒ 3x = 4x + 4

⇒ 4x – 3x = -4

⇒ x = -4

Hence the required solution are x = 3 and x = -4

Applications of Quadratic Equations in Real Life

6. \(\frac{a x+b}{a+b x}=\frac{c x+d}{c+d x}\left(a \neq b, c \neq d, x \neq-\frac{a}{b},-\frac{c}{d}\right)\)

Solution: \(\frac{a x+b}{a+b x}=\frac{c x+d}{c+d x}\left(a \neq b, c \neq d, x \neq-\frac{a}{b},-\frac{c}{d}\right)\)

or, acx + be + ad² + bdx = acx + bc² + ad + bdx

or, adx² – bcx² = ad – bc

or, x² (ad – bc) = ad – bc

or, x² = \(\frac{a d-b c}{a d-b c}\)

or, x² = 1

or, x = ±1

Hence the required solutions are x = 1, x = – 1

7. \((2 x+1)+\frac{3}{(2 x+1)}=4\left(x \neq \frac{1}{2}\right)\)

Solution: \((2 x+1)+\frac{3}{(2 x+1)}=4\left(x \neq \frac{1}{2}\right)\)

or, \(\frac{(2 x+1)^2+3}{2 x+1}=4\)

or, (2x +1)² + 3 = 4 (2x + 1)

or, (2x)² + 2.2x.1 + (1)² + 3 = 8x + 4

or, 4x² + 4x + 1 + 3 = 8x + 4

or, 4x² – 4x = 0

or, 4x (x – 1) = 0

either 4x = 0 or, x – 1 = 0

⇒ x = 0 ⇒ x =1

Hence the required solutions are x = 0 and x = 1.

8. \(\frac{x+3}{x-3}+6\left(\frac{x-3}{x+3}\right)=5(x \neq 3,-3)\)

Solution: \(\frac{x+3}{x-3}+6\left(\frac{x-3}{x+3}\right)=5(x \neq 3,-3)\)

Let \(\frac{x+3}{x-3}=a\)

∴ \(\frac{x-3}{x+3}=\frac{1}{a}\)

∴ \(a+6 \times \frac{1}{a}=5\)

or, \(\frac{a^2+6}{a}=5\)

or, a² + 6 = 5a or, a² – 5a + 6 = 0

or, a² – (2 + 3) a + 6 = 0

or, a² – 2a – 3a + 6 = 0

or, a (a- 2) – 3 (a – 2) = 0

or, (a – 2)(a – 3) = 0

∴ either a -2 = 0

⇒ a = 2

⇒ \(\frac{x+3}{x-3}=2\)

⇒ 2x-6=x + 3

⇒ 2x – x = 3 + 6

⇒ x = 9

or, a – 3 = 0

⇒ a = 3

⇒ \(\frac{x+3}{x-3}=3\)

⇒ 3x-9 = x + 3

⇒ 3x – x = 3+9

⇒ 2x = 12

⇒ x = 6

Hence the required solutions are x = 6 and x = 9.

9. \(\frac{x+1}{2}+\frac{2}{x+1}=\frac{x+1}{3}+\frac{3}{x+1}-\frac{5}{6}(x \neq-1)\)

Solution: \(\frac{x+1}{2}+\frac{2}{x+1}=\frac{x+1}{3}+\frac{3}{x+1}-\frac{5}{6}(x \neq-1)\)

or, \(\frac{x+1}{2}-\frac{x+1}{3}-\frac{3}{x+1}+\frac{2}{x+1}=-\frac{5}{6}\)

or, \(\frac{3 x+3-2 x-2}{6}-\frac{3-2}{x+1}=-\frac{5}{6}\)

or, \(\frac{x+1}{6}-\frac{1}{x+1}=-\frac{5}{6}\)

or, \(\frac{(x+1)^2-6}{6(x+1)}=-\frac{5}{6}\)

or, \(\frac{x^2+2 x+1-6}{x+1}=-5\)

or, x²+ 2x – 5 = -5x – 5

or, x² + 7x + 0

or, x(x+7) = 0

∴ either x = 0 or, x + 7 =0

⇒ x = -7

Hence the required solutions are x = -a and x = -b

10. \(\left(\frac{x+a}{x-a}\right)^2-5\left(\frac{x+a}{x-a}\right)+6=0 \quad(x \neq a)\)

Solution: \(\left(\frac{x+a}{x-a}\right)^2-5\left(\frac{x+a}{x-a}\right)+6=0 \quad(x \neq a)\)

Let \(\frac{x+a}{\dot{x}-a}=y\)

∴ y²– 5y + 6 = 0

or, y²– (2 + 3)y + 6 = 0

or, y² – 2y – 3y+ 6 = 0

or, y(y – 2) – 3(y – 2) = 0

or, (y – 2)(y – 3) = 0

either y – 2 =0

⇒ y = 2

⇒ \(\frac{x+a}{x-a}=2\)

⇒ 2x – 2a = x + a

⇒ x = 3a

or, y -3 = 0

⇒ y = 3

⇒ \(\frac{x+a}{x-a}=3\)

⇒ [putting y = \(y=\frac{x+a}{x-a}\)]

⇒ 3x – 3a = x + a

⇒ 2x = 4a

⇒ x = 2a

Hence the required solutions are x = 2a and x = 3a

11. \(\frac{1}{x}-\frac{1}{x+b}=\frac{1}{a}-\frac{1}{a+b} \quad(x \neq 0,-b)\)

Solution: \(\frac{1}{x}-\frac{1}{x+b}=\frac{1}{a}-\frac{1}{a+b} \quad(x \neq 0,-b)\)

or, \(\frac{x+b-x}{x(x+b)}=\frac{a+b-a}{a(a+b)}\)

or, \(\frac{b}{x(x+b)}=\frac{b}{a(a+b)}\)

or, \(\frac{1}{x(x+b)}=\frac{1}{a(a+b)}\)

or, x²– bx = a²+ ab

or, x² – a²+ bx – ab = 0

or, (x +a)(x-a)+b(x-a) = 0

or, (x-a)(x+a+b) = 0

∴ either x – a = 0

⇒ x – a = 0

or, x + a + b = 0

Hence the required solutions are x = a and x = -(a+b)

12. \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}(x \neq 1 ; 2,3,4)\)

Solution: \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}(x \neq 1 ; 2,3,4)\)

or, \(\left(\frac{1}{x-2}-\frac{1}{x-1}\right)+\left(\frac{1}{x-3}-\frac{1}{x-2}\right)+\left(\frac{1}{x-4}-\frac{1}{x-3}\right)=\frac{1}{6}\)

or, \(\frac{1}{x-4}-\frac{1}{x-1}=\frac{1}{6}\)

or, \(\frac{x-1-x+4}{(x-4)(x-1)}=\frac{1}{6}\)

or, \(\frac{3}{x^2-4 x-x+4}=\frac{1}{6}\)

or, \(\frac{3}{x^2-5 x+4}=\frac{1}{6}\)

or, x² – 5x + 4 = 18

or, x² – 5x + 4 – 18 = 0

or, x² – 5x – 14 = 0

or, x² – (7-2)x – 14 = 0

or, or, x² – 7x + 2x – 14 = 0

or, x(x-7) + 2(x-7) = 0

or, (x-7)(x+2) = 0

∴ either x – 7 = 0

⇒ x = 7

or, x + 2=0

⇒ x = -2

Hence the required equation are x = 7 and x = -2

13. \(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2 c}{x-c}(x \neq a, b, c)\)

Solution: \(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2 c}{x-c}(x \neq a, b, c)\)

or, \(\frac{a}{x-a}+\frac{b}{x-b}=\frac{c}{x-c}+\frac{c}{x-c}\)

or, \(\frac{a}{x-a}-\frac{c}{x-c}=\frac{c}{x-c}-\frac{b}{x-b}\)

or, \(\frac{a x-a c-c x+a c}{(x-a)(x-c)}=\frac{c x-b c-b x+b c}{(x-c)(x-b)}\)

or, \(\frac{x(a-c)}{x-a}=\frac{x(c-b)}{x-b}\) [because \(x \neq c \Rightarrow x-c \neq 0\)]

or, \(\frac{x(a-c)}{x-a}-\frac{x(c-b)}{x-b}=0\)

or, \(x\left[\frac{a-c}{x-a}-\frac{c-b}{x-b}\right]=0\)

∴ either x = 0

or, \(x\left[\frac{a-c}{x-a}-\frac{c-b}{x-b}\right]=0\)

or, \(\frac{a-c}{x-a}-\frac{c-b}{x-b}=0\)

⇒ \(\frac{(a-c)(x-b)-(x-a)(c-b)}{(x-a)(x-b)}=0\)

⇒ \(\frac{a x-c x-a b+b c-c x+a c+b x-a b}{(x-a)(x-b)}=0\)

⇒ a x+b x-2 c x+b c+c a-2 a b=0

⇒ x(a+b-2 c)=2 a b-b c-c a

⇒ \(x=\frac{2 a b-b c-c a}{a+b-2 c}\)

Hence the required solutions are x = 0 and \(x=\frac{2 a b-b c-c a}{a+b-2 c}\)

14. x²-(√3 + 2)x + 2√3 = 0

or, x²-√3x-2x + 2√3=0

or, x(x-√3)- 2(x-√3) = 0

or, (x-√3)(x-2) – 0

either x -√3 = 0 or, x – 2 = 0

⇒ x = √3 ⇒ x =2

Hence the required solution are x = √3 and x = 2

Algebra Chapter 1 Quadratic Equation In One Variable Value Of The Variable

In our real life we have to face some such problems which can be transformed into a quadratic equation in one variable that can be easily solved.

In this case, we assume a suitable part of the problem (generally the part which is to be determined) as the unknown variable.

Then we construct a quadratic equation that satisfies all the given conditions.

According to the above discussion, we determine the value of the variable by solving that quadratic equation.

The value of the variable thus obtained is the very solution of the given problem.

In the following examples, we have discussed much more about it.

Algebra Chapter 1 Quadratic Equation In One Variable Value Of The Variable

Multiple Choice Questions

Example 1. The number of roots of a quadratic equation is

1
2
3
None of these

Solution: 2. 2.

Example 2. If ax²+ bx + c = 0 be a quadratic equation, then

b≠0
c≠0
a≠0
None of these

Solution: 3. a≠0

if a = 0, then the quadratic equation is transferred into a linear one.

Example 3. The highest power of the variable of a quadratic equation is

1
2
3
None of these

Solution: 2.2

Example 4. The equation 4(5x² – 7x + 2) = 5(4x²– 6x + 3) is

Linear
Quadratic
Cubic
None of these

Solution: 1. Linear

Because, 4 (5x²-7x+ 2) = 5 (4x²– 6x + 3)

or, 20x² – 28x + 8 = 20x² – 30x+ 15

or, – 28x + 30x + 8-15 = 0

or, 2x – 7 = 0, which is a linear equation.

Example 5. Root/roots of the equation \(\frac{x^2}{x} = 6\) is/are

0
6
0 and 6
– 6

Solution: 2. 6

Because, \(\frac{x^2}{x} = 6\)= 6 implies that x ≠0

∴ x = 6

Example 6. Write which of the following are true/false:

1. (x- 3)² = x² – 6x + 9 is a quadratic equation

Solution:

1. The statement is false

Because, (x – 3)²=x² – 6x + 9

or, x² – 6x + 9 = x² – 6x + 9 or, 0 = 0.

2. 5 is the only one root of the equation x²=25

Solution: x² = 25 ⇒ x = ± 5, i.e., the roots of the given equation are 2 in number, i.e.,

x = – 5 and x = 5.

∴ The statement is false.

Example 7. Fill in the blanks

1. If a= 0 and b ≠ 0 in the equation ax² + bx + c = 0, then the equation is a _______ equation.

Solution:

1. ax²+ bx + c = 0 or, 0x² + bx + c = 0 [a = 0]

or, bx + c — 0, which is a linear equation.

∴ linear equation.

2. The roots of the equation x²=6x are _____ and ____

Solution: x² = 6x or, x² – 6x = 0

or, x (x – 6) = 0.

∴ either x = 0 or, x – 6 = 0

⇒ x = 6

∴ the roots are 0 and 6.

Algebra Chapter 1 Quadratic Equation In One Variable Value Of The Variable

Short Answer Type Questions

Example 1. Find the value of a if 1 is a root of equation x² + ax + 3 = 0.

Solution: 1 is a solution of the equation x²+ ax + 3 = 0

∴ 1² + a.1+ 3 = 0 or, 1 + a + 3 = 0 or, a = – 4.

Example 2. Determine the other root of the equation x²– (2 + b)x + 6 = 0 if one of its roots is 2.

Solution: Since 2 is a root of the equation x²-(2 + b)x + 6 = 0, we get,

2² – (2 + b).2 + 6 = 0 or, 4 – 4 – 2b + 6 = 0

or, 2b = 6 or, b = \(\frac{6}{2}\) = 3.

Hence, the other root is 3.

Aliter: Product of the roots = \(\frac{6}{1}\) =6

One of the roots is 2.

∴ the other root is \(\frac{6}{2}\) = 3

Example 3. If 2 is a root of the equation 2x²+ kx + 4 = 0, then what is the value of the other root?

Solution: Product of the roots = \(\frac{4}{2}\) = 2 [by the formula, product of the roots = \(\frac{c}{a}\)]

One of the roots = 2, the other root = \(\frac{2}{2}\) = 1.

Example 4. The difference of a proper fraction and its reciprocal is \(\frac{9}{20}\) Then find the equation.

Solution: Let the proper fraction be \(\frac{1}{x}\) ∴ the reciprocal =x

As per question, \(\frac{6}{2}\) – x = \(\frac{9}{20}\)

Hence, the required equation is \(\frac{1}{x}\) – x = \(\frac{9}{20}\)

Example 5. If – 5 and – 7 be two roots of the equation ax²+ bx + 35 = 0, then find the values of a and b.

Solution: ax²+ bx + 35 – 0

or, a.(- 5)² + b. (-5) + 35 = 0 [(-5) is a root ]

or, 25a – 5b + 35 = 0

. or, 5a – b + 7 = 0….. (1)

Also, ax² + bx + 35 = 0

or, a x (-7)² + b x (-7) + 35 = 0 [(-7) is a root ]

or, 49a – 7b + 35 = 0

or, 7a – b + 5 = 0…..(2)

Now, subtracting (1) from (2), we get, 2a – 2 = 0 or, 2a = 2 or, a = 1.

Putting a = 1 in (1) we get, 5 x 1-b + 7 = 0

⇒ 5 -b + 7 = 0

⇒ 12 – b = 0 ⇒ b = 12.

Hence a = 1 and b = 12.

Algebra Chapter 1 Quadratic Equation In One Variable Value Of The Variable Long Answer Type Questions

Example 1. The difference of two positive whole numbers is 3 and the sum of their squares is 117. Find the two numbers.

Solution:

Given:

The difference of two positive whole numbers is 3 and the sum of their squares is 117.

Let one of the numbers = x,

∴ the other number = (x – 3) or (x + 3)

As per question, x² + (x – 3)² = 117

or, x² + x² – 6x + 9 = 177

or, 2x² – 6x + 9 – 117 = 0

or, 2x² – 6x – 108 = 0

or, x² – 3x – 54 = 0

or, x² – (9 – 6) x – 54 = 0

or, x² – 9x + 6x – 54 = 0

or, x (x – 9) + 6 (x – 9) = 0

or, (x – 9)(x + 6) = 0

∴ either x – 9 = 0 or,x + 6 = 0

⇒ x = 9 or, x = -6

∴ x is positive,

∴ x = – 6.

and x² + (x + 3)² = 117

or, x² + x² + 6x + 9 = 117

or, 2x² + 6x + 9 – 117 = 0

or, 2x² + 6x – 108 = 0

or, x² + 3x – 54 = 0

or, x² + (9 – 6) x – 54 = 0

or, x² + 9x – 6x – 54 = 0

or, x (x + 9) – 6 (x + 9) = 0

or, (x + 9)(x – 6) = 0

either x + 9 = 0 or, x-6 = 0

⇒ x = – 9 or, x = 6.

∴ x is positive, x ≠ – 9.

Hence the required two numbers are 6 and 9

Example 2. The base of a triangle is 18m more than double of its height. If the area of the triangle be 360 sq-, then find the height of the triangle.

Solution:

Given:

The base of a triangle is 18m more than double of its height. If the area of the triangle be 360 sq

Let the height of the triangle = xm,

∴ its base = (2x + 18 ) m

∴ area of the triangle = \(\frac{1}{2}\) X (2x + 18) X x sq-m

= (x² + 9x) sq-m.

As per question, x² + 9x = 360

or, x² + (24 – 15)x – 360 = 0

or, x² + 24x – 15x – 360 = 0

or, x (x + 24) – 15 (x + 24) = 0

or, (x + 24)(x – 15) = 0

∴ either x + 24 = 0 or, x – 15 = 0

⇒ x = -24 ⇒ x = 15

But height can not be negative, ∴x ≠ -24.

∴ x = 15

Hence the height of the triangle is 15 m.

Graphing Quadratic Functions

Example 3. If 5 times of a positive whole number is 3 less than 2 times of its square, then find the number.

Solution:

Given:

If 5 times of a positive whole number is 3 less than 2 times of its square

Let the number = x (x > 0)

As per questions, 5x = 2x² – 3

or, 2x² – 5x – 3 = 0

or, 2x²– (6- 1)x – 3 = 0

or, 2x² -6x + x- 3 = 0

or, 2x (x – 3) + 1 (x – 3) = 0

or, (x-3)(2x+1) = 0

∴ either x-3 = 0 or, 2x+1 = 0

⇒ x = 3 or, 2x = -1

or, x= –\(\frac{1}{2}\)

Since the number is a whole number,

x = 3

Hence the required positive whole number is 3.

Example 4. The distance between two places is 200 km. The time taken by motor car to travel from one place to another is less by 2 hrs than the time taken by a zeep car. If the speed of the motor car is 5 km/hr more than the speed of the zeep car, then find the speed of the motor car.

Solution:

Given:

The distance between two places is 200 km. The time taken by motor car to travel from one place to another is less by 2 hrs than the time taken by a zeep car. If the speed of the motor car is 5 km/hr more than the speed of the zeep car

Let the speed of the motor car be x km/hr.

∴ Speed of the zeep car = (x + 5) km/hr.

∴ The time taken by the motor car to travel 200 km is hr and by zeep car it is \(\frac{200}{x}\) hr

and by zeep car it is \(\frac{200}{x+5}\)

As per the question, \(\frac{200}{x}\)– \(\frac{200}{x+5}\) = 2

or, \(200\left(\frac{1}{x}-\frac{1}{x+5}\right)\) =2

or, \(200\left\{\frac{x+5-x}{x(x+5)}\right\}=2\) =2

or, \(200 \times \frac{5}{x^2+5 x} \) =2

or, 2(x²+5x) = 1000

or, x² + (25 – 20)x – 500

or, x² + 25x – 20x – 500 = 0

or, x(x+ 25) – 20 (x + 25) = 0

or, (x + 25)(x – 20) = 0

∴ either x+25 =0 or, x- 20 = 0

But speed can not be negative,∴ x = 20

∴ The required speed of the motor car is 20 km/hr.

Example 5. The tens’ digit of a two-digit number is less by 3 than the unit digit. If the product of the two digits is subtracted from the number, the result is 15. Find the unit digit of the number.

Solution:

Given:

The tens’ digit of a two-digit number is less by 3 than the unit digit. If the product of the two digits is subtracted from the number, the result is 15.

Let the unit digit be x

∴ The tens’ digit is (x- 3)

∴ the number = 10 (x – 3) + x.

= 10x – 30 + x = 11x – 30.

As per questions, (11x- 30) – x (x – 3) = 15.

or, 11x- 30 – x² + 3x = 15

or, x² -14x + 45 = 0

or, x² – (5 + 9)x+ 45 = 0

or, x² – 5x – 9x + 45 = 0

or, x (x – 5) – 9 (x-5) = 0

or, (x – 5)(x – 9) = 0

∴ either x-5 = 0 or, x – 9 = 0

⇒ x = 5 or, x = 9

Hence the unit digit is 5 or 9

Example 6. There are two pipes in the water reservoir of your school. Two pipes together take 11 \(\frac{1}{9}\) minutes to fill the reservoir. If the two pipes are opened separately, then one pipe would take 5 minutes more time than the other pipe. Calculate the time taken to fill the reservoir separately by each of the pipes.

Solution:

Given :

There are two pipes in the water reservoir of your school. Two pipes together take 11 \(\frac{1}{9}\) minutes to fill the reservoir. If the two pipes are opened separately, then one pipe would take 5 minutes more time than the other pipe.

Let the first pipe can fill the reservoir separately in x minutes

∴ the second pipe can fill the reservoir separately in (x + 5) minutes.

Two pipes together can fill the reservoir in 1 minute

\(\left(\frac{1}{x}+\frac{1}{x+5}\right)\) part

= \(\frac{x+5+x}{x(x+5)}\) part

= \(\frac{2 x+5}{x^2+5 x}\) part of the reservoir.

So, the Two pipes together can fill the reservoir in 1 minute \(\frac{x^2+5 x}{2 x+5}\) part of it.

As per question, \(\frac{x^2+5 x}{2 x+5}\) = 11\(\frac{1}{9}\)

or, \(\frac{x^2+5 x}{2 x+5}=\frac{100}{9}\)

or, 9x² +45x= 200x +500

or, 9x² – 155x – 500 = 0

or, 9x²-(180-25)x-500 = 0

or, 9x² – 180x + 25x – 500 = 0

or, 9x (x – 20) + 25 (x – 20) = 0

or, (x – 20)(9x + 25) = 0

∴ either x-20=0 or, 9x + 25 = 0

⇒ x =20 or, 9x = – 25

⇒ x =20 or, x= –\(\frac{25}{9}\)

But the value of x can not be negative,

∴ x ≠ –\(\frac{25}{9}\)

∴ x = 20

Hence the first pipe in 20 minutes and the second pipe in 25 minutes can fill in the reservoir separately.‘

Example 7. Laxmi and Kartik together complete a work in 4 days. If they work separately, then the time taken by Kartik would be 6 days more than the time taken by Laxmi. Calculate the time taken by Laxmi alone to complete the work.

Solution:

Given:

Laxmi and Kartik together complete a work in 4 days. If they work separately, then the time taken by Kartik would be 6 days more than the time taken by Laxmi.

Let Laxmi alone complete the work in x

∴ Kartik alone can complete the work in (x + 6) days.

They together complete \(\left(\frac{1}{x}+\frac{1}{x+6}\right)\) part of the work in 1 day, i.e.,

they complete \(\frac{x+6+x}{x(x+6)}\) part

= \(\frac{2 x+6}{x^2+6 x}\) parts of the work in 1 day

∴ They together complete the work in \(\frac{x^2+6 x}{2x+6}\) days.

As per question, \(\frac{x^2+6 x}{2x+6}\) = 4

or, x² + 6x = 8x + 24

or, x² – 2x – 24 = 0

or, x² – (6 – 4) x – 24 = 0

or, x² – 6x + 4x – 24 = 0

or, x (x-6)+ 4 (x-6) = 0

or, (x – 6)(x + 4) = 0

∴ either x – 6 = 0 or, x + 4 = 0

⇒ x = 6 or, x = – 4.

But the number of days can not be negative, ∴ x ≠ – 4, ∴ x = 6.

Hence, Laxmi alone can complete the work in 6 days.

Example 8. If the price of 1 dozen pen is reduced by ₹6, then 3 more pens will be got at ₹30. Before the reduction of price, calculate the price of 1 dozen pen.

Solution: Let the price of 1 dozen pen at present is ₹ x.

∴ in ₹ 30, at present \(\frac{12}{x}\) x 30pen = \(\frac{360}{x}\) pen are got.

If the price is reduced by ₹6 per dozen, then it becomes ₹(x – 6).

Then for ₹30 we get, \(\frac{12}{x-6}\)x30 pens = \(\frac{360}{x-6}\) pens.

As per question, \(\frac{360}{x-6}\) – \(\frac{360}{x}\) = 3

or, \(360\left(\frac{1}{x-6}-\frac{1}{x}\right)\)=3

or, \(120\left\{\frac{x-x+6}{(x-6) x}\right\}\)=1

or, \(120\left(\frac{6}{x^2-6 x}\right)=1\)

or, \(\frac{720}{x^2-6 x}=1\)

or, x²– 6x – 720 = 0

or, x²– (30 – 24) x – 720 = 0

or, x²– 30x + 24x – 720 = 0

or, x (x – 30) + 24(x – 30) = 0

or, (x – 30)(x + 24) = 0

∴ either x – 30 = 0 or, x + 24 = 0

⇒ x = 30 or, x = – 24.

But price of pens can not be negative,

∴ x≠- 24, x = 30.

Hence, before reduction of prices, the price of 1 dozen pen was ₹30.

Algebra Chapter 1 Quadratic Equation In One Variable Determination Of Roots Of Quadratic Equations By Expressing It In Perfect Squares Or Hindu Method

You have already known that the standard form of a quadratic equation is ax² + bx + c = 0, where a, b, c are reals and a≠0.

To find the roots of this equation, we have taken the method of factorisation in the previous section.

In this section, to determine the roots- of this equation, we shall first express the equation as a perfect square.

Then we shall determine the roots

Now, ax² + bx + c = 0,

or, \(x^2+\frac{b}{a} x+\frac{c}{a}=0\) [Dividing by a, a≠0]

or, \(x^2+2 \cdot \frac{b}{2 a} \cdot x+\frac{c}{a}=0\)

or, \(x^2+2 \cdot x \cdot \frac{b}{2 a}+\left(\frac{b}{2 a}\right)^2+\frac{c}{a}=0+\left(\frac{b}{2 a}\right)^2\)

[Adding \(\left(\frac{b}{2 a}\right)^2\) in both the sides]

or, \(\left(x+\frac{b}{2 a}\right)^2+\frac{c}{a}=\frac{b^2}{4 a^2}\)

or, \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

or, \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 c a}{4 a^2}\) …(1)

∴ expressing the quadratic equation in perfect square we get,

\(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 c a}{4 a^2}\)

We can write from this equation \(x+\frac{b}{2 a}=\sqrt{\frac{b^2-4 a c}{4 a^2}}\)

[squaring on both sides]

or, \( x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)

or, \( x=-\frac{b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}or, \)

or, \( x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

We know that a quadratic equation has at least two roots.

So, the roots are \(\frac{-b+\sqrt{b^2-4 a c}}{2 a} \text { and } \frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

Hence the roots of a quadratic equation ax² + bx + c = 0, where a, b, c are reals and a ≠ 0, are

\(\frac{-b+\sqrt{b^2-4 a c}}{2 a} \text { and } \frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

It is known as Sreedhar Acharya’s formula of finding the roots of a quadratic equation.

It is also known as Hindu method of finding the roots of a quadratic equation.

According to the name of well-known Indian mathematician Sreedhar Acharya, this formula is called Sreedhar Acharya’s formula.

We can easily determine the roots of any quadratic equation with the help of this formula.

Example 1. Determine the roots of equation 3x² + 2x-1=0applying Sreedhar Acharya’s formula.

Solution: The given equation is 3x²+ 2x – 1 = 0.

Comparing with the standard equation ax²+ bx + c = 0, we get , a = 3, b = 2 and c = – 1.

∴ The required roots are

\(\frac{-2 \pm \sqrt{(2)^2-4 \cdot 3 \cdot(-1)}}{23}=\frac{-2 \pm \sqrt{4+12}}{6}\)

= \(\frac{-2 \pm \sqrt{16}}{6}=\frac{-2 \pm 4}{6}=\frac{1}{3}\) [taking +sign]

and (-1) (taking – sign).

Hence the roots are : x = \(\frac{1}{3}\) and x = -1

Example 2. Expressing in a perfect square find the roots of the quadratic equation 10x² – x – 3 = 0

Solution: Given that 10x² – x – 3 = 0

or, \(x^2-\frac{x}{10}-\frac{3}{10}=0\)

[Note: Generally division is made by non-zero co-efficient of x²]

or, \(x^2-2 \cdot x \cdot \frac{1}{20}+\left(\frac{1}{20}\right)^2-\frac{3}{10}=0+\left(\frac{1}{20}\right)^2\)

[ Adding \(\left(\frac{1}{20}\right)^2\) on both sides]

or, \(\left(x-\frac{1}{20}\right)^2=\frac{1}{400}+\frac{3}{10}\)

or, \(\left(x-\frac{1}{20}\right)^2=\frac{1+120}{400}\)

or, \(\left(x-\frac{1}{20}\right)^2=\frac{121}{400}\)

∴ Expressing the given quadratic equation a perfect square we get,

\(x-\frac{1}{20}=\sqrt{\frac{121}{400}}\)

(taking square root of both sides)

or, x-\(\frac{1}{20}\) = ± \(\frac{11}{20}\)

or, x = \(\frac{1}{20}\) ± \(\frac{11}{20}\)

or, \(x=\frac{1 \pm 11}{20}\)

∴ The required roots are

\(x=\frac{1+11}{20}\) = \(x=\frac{3}{5}\)

and \(x=\frac{1-11}{20}\) = – \(x=\frac{1}{2}\)

Algebra Chapter 1 Quadratic Equation In One Variable Multiple Choice Questions

Example 1. The perfect square form of the quadratic equation ax²+ bx+ c = 0 (a ≠ 0) is

\(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)
\(\left(x-\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)
\(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a}\)
\(\left(x-\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a}\)

Solution: Given that ax²+ bx+ c = 0 (a ≠ 0)

or, \(\quad x^2+\frac{b}{a} x+\frac{c}{a}=0\)

or, \(x^2+2 \cdot x \cdot \frac{b}{2 a}+\left(\frac{b}{2 a}\right)^2+\frac{c}{a}=\left(\frac{b}{2 a}\right)^2\)

or, \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

or, \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

Hence the required perfect square form is \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

∴ 2. \(\left(x-\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\) is correct

Example 2. The perfect square form of the quadratic equation 5x² + 23x + 120 = 0 is

\(\left(x-\frac{23}{10}\right)^2=\left(\frac{17}{10}\right)^2\)
\(\left(x+\frac{23}{10}\right)^2=\left(\frac{17}{10}\right)^2\)
\(\left(x-\frac{23}{10}\right)^2=\left(\frac{19}{10}\right)^2\)
\(\left(x+\frac{23}{10}\right)^2=\left(\frac{19}{10}\right)^2\)

Solution: Given that 5x² + 23x + 120 = 0

or, \(x^2+\frac{23}{5} x+\frac{12}{5}=0\) [Dividing by 5 ]

or, \(x^2+2: x \cdot \frac{23}{10}+\left(\frac{23}{10}\right)^2+\frac{12}{5}=\left(\frac{23}{10}\right)^2\)

[Adding \(\left(\frac{23}{10}\right)^2\) in both the sides]

or, \(\left(x+\frac{23}{10}\right)^2=\frac{529}{100}-\frac{12}{5}\)

or, \(\left(x+\frac{23}{10}\right)^2=\frac{529-240}{100}\)

or, \(\left(x+\frac{23}{10}\right)^2=\frac{289}{100}\)

or, \(\left(x+\frac{23}{10}\right)^2=\left(\frac{17}{10}\right)^2\)

Hence the required perfect square form is \(\left(x+\frac{23}{10}\right)^2=\left(\frac{17}{10}\right)^2\)

∴ 2. \(\left(x+\frac{23}{10}\right)^2=\left(\frac{17}{10}\right)^2\)

Example 3. If the equation (x- 2)(x + 4) + 9 = 0 be expressed in the form ax² + bx + c = 0 (a ≠ 0), then the value of b is

Solution: Given that (x – 2)(x + 4) + 9 = 0

or, x² – 2x + 4x-8 + 9 = 0

or, x² + 2x + 1 = 0

Comparing with the equation ax² + bx + c = 0 we get, b = 2.

Hence b = 2.

∴ 3.2 is correct.

Example 4. If the equation (4x – 3)² – 2 (x + 3) = 0 is expressed in the form ax² + bx + c = 0 (a ≠ 0), then the value of a is

– 26
26
– 16
16

Solution: Given that (4x- 3)²-2 (x + 3) = 0

or, (4x)² – 2.4x.3 + (3)² – 2x – 6 = 0

or, 16x² – 24x + 9- 2x-6 = 0

or, 16x²-26x + 3 = 0 .

Comparing with the equation ax² + bx + c = 0 we get, a = 16.

Hence the required value of a is 16.

∴ 4.16 is correct.

Example 5. Write true or false

1. If the equation (x+2)(x-4)+10= 0 be expressed in the form ax² + bx + c = 0 then the value of c is 2.

Solution: True

Since (x + 2) (x- 4) + 10 = 0

⇒ x² + 2x – 4x – 8 + 10 = 0 .

⇒ x²-2x + 2 = 0

Comparing with ax² + bx + c = 0 we get c = 2.

Hence the statement is true.

2. The perfect square form of the quadratic equation 3x² + 17x + 11 = 0 is

\(\left(x+\frac{17}{6}\right)^2=\left(\frac{157}{6}\right)^2\)

Solution: False 3X²+ 17x +11=0

or, \(x^2+\frac{17}{3} x+\frac{11}{3}=0\) [Dividing by 3]

or, \((x)^2+2 \cdot x \cdot \frac{17}{6}+\left(\frac{17}{6}\right)^2=\frac{11}{3}+\left(\frac{17}{6}\right)^2\)

or, \(\left(x+\frac{17}{6}\right)^2=-\frac{11}{3}+\frac{289}{36}\)

or, \(\left(x+\frac{17}{6}\right)^2=\frac{-132+289}{36}\)

or, \(\left(x+\frac{17}{6}\right)^2=\frac{157}{36}\)

or, \(\left(x+\frac{17}{6}\right)^2=\left(\frac{\sqrt{157}}{6}\right)^2\)

Hence the statement is false.

Example 6. Fill in the blanks

1. The value of k is x= \(\frac{k \pm 7}{2}\) when the quadratic equation x²+x-12 = 0 is solved by Sreedhar Acharya’s formula is ____

Solution: k = 1

Since the given equation x²+x-12 = 0

∴ \( x =\frac{-1 \pm \sqrt{(a)^2-4.1 \times(-12)}}{2.1}\)

= \(\frac{-1 \pm \sqrt{1+48}}{2}\)

= \(\frac{-1 \pm \sqrt{49}}{2}=\frac{-1 \pm 7}{2}\) ….(1)

∴ Comparing (1) with x= \(\frac{k \pm 7}{2}\) we get k=-1

2. If the equation (3x – 1)²+ 2 (x – 3) = 0 is expressed in the form ax² + bx + c = 0 (a ≠ 0), then the value of a is _______

Solution: – 4 ; since the given equation is (3x- 1)² + 2 (x – 3) = 0

or, (3x)² – 2.3×1 + (1)² + 2x – 6 = 0

or, 9x² – 6x + 1 + 2x – 6 = 0

or, 9x² – 4x – 5 = 0 ….(2)

Comparing (2) with ax² + bx + c = 0 we get, b = – 4.

Algebra Chapter 1 Quadratic Equation In One Variable Short Answer Type Questions

Example 1. Explain whether Sreedhar Acharya’s formula is applicable or not in solving the equation 4x² + (2x-1)(2x + 1) = 4x (2x – 1)

Solution: Given that 4x²+ (2x – 1)(2x + 1) = 4x (2x – 1)

4x² + (2x)²-(1)² = 8x²-4x

or, 4x² + 4x² – 1 = 8x²– 4x

or, 8x²– 1 = 8x² – 4x

or, 8x²– 1 – 8x² + 4x = 0

or, 4x – 1 = 0, which is a linear equation.

Since the given equation is not a quadratic one, Sreedhar Acharya’s formula is not applicable in solving the equation.

Example 2. By applying Sreedhar Acharya’s formula what type of equations can we solve?

Solution: Quadratic equations in one variable.

Example 3. What is the value of kif x = \(\frac{k+2}{10}\) when the quadratic equation 5x² + 2x – 7 = 0 is solved by Sreedhar Acharya’s formula?

Solution: Comparing the given equation 5x²+ 2x – 7 = 0 with ax² + bx + c = 0 (a ≠0)

we get, a = 5, b = 2 and c = – 7

∴ According to Sreedhar Acharya s formula,

\(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-2 \pm \sqrt{(2)^2-4 \times 5 \times(-7)}}{2 \times 5}\)

= \(\frac{-2 \pm \sqrt{4+140}}{10}=\frac{-2 \pm \sqrt{144}}{10}=\frac{-2 \pm 12}{10}\)

= \(\frac{-2+12}{10} \text { (taking }+ \text { sign) and } \frac{-2-12}{10}\) (taking – sign)

Now given that x = \(\frac{k+12}{2} \cdots \cdots \cdots(1)\)

Also x= \(\frac{-2+12}{10} \cdots \cdots \cdots(2)\)

∴ Comparing (1) and (2), we get, k = – 2.

Hence the required value of k = – 2.

Example 4. Express the equation cx²+ ax + b = 0, c ≠0 in the form of a perfect square.

Solution: Given that cx² + ax + b = 0

or, \(x^2+\frac{a}{c} x+\frac{b}{c}=0\) [Dividing by c, c ≠ 0]

or, \(x^2+2 \cdot x \cdot \frac{a}{2 c}+\left(\frac{a}{2 c}\right)^2+\frac{b}{c}=0+\left(\frac{a}{2 c}\right)^2\)

or, \(\left(x+\frac{a}{2 c}\right)^2=\frac{a^2}{4 c^2}-\frac{b}{c}\)

or, \(\left(x+\frac{a}{2 c}\right)^2=\frac{a^2-4 b c}{4 c^2}\)

Hence the perfect square form is : \(\left(x+\frac{a}{2 c}\right)^2=\frac{a^2-4 b c}{4 c^2}\)

Algebra Chapter 1 Quadratic Equation In One Variable Long Answer Type Questions

Example 1. Multiplying LHS and RHS of the quadratic equation 5X²+ 23x + 12 = 0 by 5, find the roots of the equation using perfect square method.

Solution: Given that 5x² + 23x +12 = 0

or, 25x² + 115x + 60 = 0 [Multiplying by 5]

or,\((5 x)^2+2.5 x \cdot \frac{23}{2}+\left(\frac{23}{2}\right)^2=-60+\left(\frac{23}{2}\right)^2\)

[Adding \(\left(\frac{23}{2}\right)^2\) in both the sides]

or, \(\left(5 x+\frac{23}{2}\right)^2=-60+\frac{529}{4}\)

or, \(\left(5 x+\frac{23}{2}\right)^2 =\frac{-240+529}{4}\)

or, \(\left(5 x+\frac{23}{2}\right)^2 =\frac{289}{4}\)

or, \(\left(5 x+\frac{23}{2}\right)^2 =\left(\frac{17}{2}\right)^2\)

or, \(5 x+\frac{23}{2}= \pm \frac{17}{2}\)

or, \(5 x+\frac{23}{2}=\frac{17}{2}\) (taking + sign) and \(5 x+\frac{23}{2}=-\frac{17}{2}\) (taking – sign)

or, \(5 x=\frac{17}{2}-\frac{23}{2}\) or, \(5 x=-\frac{17}{2}-\frac{23}{2}\)

or, \(5 x=\frac{17-23}{2}\) or, \(5 x=\frac{-17-23}{2}\)

or, \(5 x=\frac{-6}{2}\) or, \(5 x=\frac{-40}{2}\)

or, 5x = -3 or, 5x = -20

or, x = – \(\frac{3}{5}\) or, x = -4

Hence the roots of the quadratic equation 5x² + 23x + 12 = 0 are – \(\frac{3}{5}\) and (-4).

Example 2. Solve the following equations using Sreedhar Acharya’s formula :

1. (x-2)(x + 4) + 9 = 0

Solution: (x-2)(x + 4) + 9 = 0

or, x² – 2x + 4x – 8 + 9 = 0

or, x² + 2x +1 = 0 ….. (1)

Comparing (1) with ax² + bx + c = 0, (a≠ 0) we get, a = 1,6 = 2 and c – 1

∴ according to Sreedhar Acharya’s formula we get,

= \( x=\frac{-2 \pm \sqrt{(2)^2-4.1 .1}}{2 \times 1}\)

or, \( x=\frac{-2 \pm \sqrt{4-4}}{2}\)

or, \( x=\frac{-2 \pm 0}{2}=\frac{-2}{2}=-1\)

Hence the required solution is x = – 1.

2. 10x²-x-3= 0

Solution: Given that 10x²-x-3= 0 …… (1)

Comparing (1) with ax² + bx + c – 0 we get, a = 10, b = – 1 and c = – 3.

∴ by Sreedhar Acharya s formula we get,

\(x=\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 10 \times(-3)}}{2 \times 10}\)

= \(\frac{1 \pm \sqrt{121}}{20}=\frac{1 \pm 11}{20}\)

=\(\frac{1+11}{20}\) [taking +sign] and \(\frac{1-11}{20}\) (taking }-sign)

= \(\frac{12}{20} \text { and } \frac{-10}{20}\)

= \(\frac{3}{5} \text { and }-\frac{1}{2}\)

Hence the required solutions are x = \(\frac{3}{5}\) and \(-\frac{1}{2}\)

3. 3x²+2x-1 =0

Solution: Comparing the given equation 3x²+2x-1 = 0 with a²x+bx +c = 0,(a 0) we get a = 3, b=2 and c = -1

By sreedhar acharya’s formula we get

\(x=\frac{-2 \pm \sqrt{(2)^2-4 \times 3 \times(-1)}}{2 \times 3}\)

=\(\frac{-2 \pm \sqrt{4+12}}{6}=\frac{-2 \pm \sqrt{16}}{6}=\frac{-2 \pm 4}{6}\)

= \(\frac{-2+4}{6}\) (taking +sign) and \(\frac{-2-4}{6}\)(taking – sign)

= \(\frac{2}{6} \text { and } \frac{-6}{6}\)

= \(\frac{1}{3} \text { and }(-1)\)

Hence the required solutions are x= \(\frac{1}{3} \text { and }(-1)\)

4. The given equation 25x²-30x+7=0 is compared with ax²+bx+c = 0,(a ≠0) and

Solution: we get a = 25, b = -30 and c= 7

∴ by Sreedhar Acharya’s formula we get,

x= \(\frac{-(-30)\pm\sqrt{(-30)^2-4\times 25 \times 7}}{2 \times \cdot 25}\)

or, \(x=\frac{30 \pm \sqrt{900-700}}{50}\)

or, \(x=\frac{30 \pm \sqrt{200}}{50}\)

or, \(x=\frac{30 \pm 10 \sqrt{2}}{50}\)

or, \(x=\frac{10(3 \pm \sqrt{2})}{50}\)

or, \(x=\frac{3+\sqrt{2}}{5}\) (taking + sign) and \(x=\frac{3-\sqrt{2}}{5}\) (taking – sign)

Hence the required solution are \(x=\frac{3+\sqrt{2}}{5}\) and \(x=\frac{3-\sqrt{2}}{5}\)

Example 3. Express the following mathematical problems in quadratic equations in one variable and solve them using Sreedhar Acharya’s formula

1. Adhir has drawn a right-angled triangle whose length of hypotenuse is 6 cm more than twice of the shortest side. If the length of the third side is 2 cm less than the length of the hypotenuse, then calculate the lengths of the three sides of the right-angled triangle drawn by Adhir.

Solution:

Given:

Adhir has drawn a right-angled triangle whose length of hypotenuse is 6 cm more than twice of the shortest side. If the length of the third side is 2 cm less than the length of the hypotenuse

Let the length of the smallest side of the right-angled triangle be x cm.

∴ The length of the hypotenuse = (2x + 6) cm

and the length of the third side = (2x + 6 – 2) cm = (2x + 4) cm.

By Pythagoras’ theorem we get, (2x + 6)² = x² + (2x + 4)²

or, (2x + 6)² – (2x + 4)² = x²

or, (2x + 6 + 2x + 4)(2x + 6 – 2x – 4) = x²

or, (4x + 10) x 2 = x²

or, 8x + 20 = x²

or, x² – 8x – 20 = 0 …… (1)

Comparing (1) with ax² + bx + c = 0, (a ≠ 0) we get, a= 1, 6 = – 8 and c = -20.

∴ by Sreedhar Acharya’s formula we get,

x= \(\frac{-(-8) \pm \sqrt{(-8)^2-4 \times 1 \times(-20)}}{2 \times 1}\)

or, \(x=\frac{8 \pm \sqrt{64+80}}{2}\)

or, \(x=\frac{8 \pm \sqrt{144}}{2}\)

or, \(x=\frac{8 \pm 12}{2}\)

∴ \(x=\frac{8+12}{2}\) (taking + sign) and x= \(\frac{8-12}{2}\)(taking – sign)

or, \(x=\frac{20}{2}\)

or, \(x=\frac{-4}{2}\)

or, x = 10 or, x = -2

But the length of a side can never be negative, x ≠ – 2, i.e., x = 10.

∴ Hypotenuse = (2x + 6) cm – (2 x 10 + 6) cm = 26 cm.

Third side = (2x + 4) cm = (2 x 10 + 4) cm = 24 cm.

Hence the sides of the triangle are 10 cm, 24 cm, and 26 cm.

2. If a two-digit positive number is multiplied by its unit digit, then the product is 189 and if the tens’ digit is twice of the unit digit, then calculate the unit digit.

Solution:

Given

If a two-digit positive number is multiplied by its unit digit, then the product is 189 and if the tens’ digit is twice of the unit digit

Let the unit digit = x, ∴ tens’s digit = 2x.

∴ the number = 10 x 2x + x = 20x + x = 21x.

As per question, 21x X x = 189, or, 21x² = 189

or,x² = \(\frac{189}{21}\) or x² = 9 or, x = ± 3.

But the value of x can not be negative, ∴x= 3

Hence the required unit digit = 3

3. There is a squared park in the locality of Prodipbabu. The area of a rectangular park is 78 sq-m less than the twice of the area of that square-shaped park whose length is 5m more than the length of the side of that park and the breadth is 3 m less than the length of the side of that park. Calculate the length of the side of the square-shaped park.

Solution:

Given:

There is a squared park in the locality of Prodipbabu. The area of a rectangular park is 78 sq-m less than the twice of the area of that square-shaped park whose length is 5m more than the length of the side of that park and the breadth is 3 m less than the length of the side of that park.

Let the length of the side of the square-shaped park be x cm.

∴ Area of park = x²sq-m.

As per question, (x+5)(x-3) = 2x²-78

or, x²+5x-3x-15 = 2x²-78

or, x²+2x-15 = 2x²-78

or, 2x²-78 – x²+5x-3x-15 = 0

or, x² -2x -63 = 0 …(1)

Comparing(1) with ax²+bx+c=0 we get a = 1,b=-2 and c=-63

∴ by Sreedhar Acharya’s formula we get

x=\(\frac{-(-2) \pm \sqrt{(-2)^2-4 \times 1 \times-63}}{2 \times 1}\)

= \(\frac{2 \pm \sqrt{4+252}}{2}=\frac{2 \pm \sqrt{256}}{2}=\frac{2 \cdot \pm 16}{2}\)

= \(\frac{2+16}{2}\)[taking + sign] and \(x=\frac{2-16}{2}\)(taking – sign)

= 9 and x = -7

Both the length of a side can never be negative

∴ x ≠-7 ,x = 9

4. Joshep and Kuntal work in a factory. Joshep takes 5 minutes less time than Kuntal to make a product. Joshep makes 6 product more than Kuntal while working for 6 hours. Calculate the number of products Kuntal makes during that time.

Solution:

Given:

Joshep and Kuntal work in a factory. Joshep takes 5 minutes less time than Kuntal to make a product. Joshep makes 6 product more than Kuntal while working for 6 hours.

Let Kuntal makes x products in 6

∴ Joshep makes (x + 6) products in 6 hours.

∴ Kuntal makes 1 product in \(\frac{6}{x}\) hours = \(\frac{6}{x}\)x 60 minutes = \(\frac{360}{x}\) minutes.

and Joshep makes 1 product in \(\frac{6}{x+6}\) hours

= \(\frac{6}{x+6}\) X 60 minutes

= \(\frac{360}{x+6}\) minutes

As per question, \(\frac{360}{x}-{360}{x+6}\)

or, \(360\left(\frac{1}{x}-\frac{1}{x+6}\right)=5\)

or, \(72\left(\frac{1}{x}-\frac{1}{x+6}\right)=1\)

or, \(\frac{x+6-x}{x(x+6)}=\frac{1}{72}\)

or, \(\frac{6}{x^2+6 x}=\frac{1}{72}\)

or, x²+6x=432

or, x²+6x-432 = 0….(1)

Comparing (1) with ax² + bx + c = 0, (a ≠ 0) we get, a = 1, 6 = 6 and c = – 432.

∴ by Sreedhar Acharya’s formula we get

\( x=\frac{-6 \pm \sqrt{(6)^2-4 \times 1 \times(-432)}}{2 \times 1}\)

x= \( \frac{-6 \pm \sqrt{36+1728}}{2}\)

x= \( \frac{-6 \pm \sqrt{1764}}{2}\)

or, \( x=\frac{-6 \pm 42}{2}\)

∴ x = \(=\frac{-6+42}{2}\) (taking }+ sign) and x= \( \frac{-6-42}{2}\) (taking – sign) }

=\(\frac{36}{2}\) and \(x=\frac{-48}{2}\)

= 18 and x = -24

But the number of products can not be negative, x = 18.

Hence Kuntal makes 18 products in 6 hours.

5. The speed of a boat in still water is 8 km/hr. If the boat can go 15 kms downstream and 22 km upstream in 5 hours, then calculate the speed of the stream.

Solution:

Given

The speed of a boat in still water is 8 km/hr. If the boat can go 15 kms downstream and 22 km upstream in 5 hours

Let the speed of the stream be x km/hr.

In still water the speed of the boat is 8 km/hr.

∴ the speed of boat in down-stream = (8 + x) km / hr and in up-stream = (8 – x) km / hr.

As per question, \(\frac{15}{8+x}+\frac{22}{8-x}=5\)

or, \(\frac{15(8-x)+22(8+x)}{(8+x)(8-x)}=5\)

or, \(\frac{120-15 x+176+22 x}{64-x^2}=5\)

or, 7x + 296 = 320 – 5X²

or, 5x² + 7x + 296 – 320 = 0

or, 5x² + 7x – 24 = 0 …. (1)

Comparing (1) with ax²+bx+c = 0 we get, a = 5, b = 7, c = -24

∴ x=\(\frac{-7 \pm \sqrt{(7)^2-4 \times 5 \times(-24)}}{2 \times 5}\)

= \(\frac{-7 \pm \sqrt{49+480}}{10}=\frac{-7 \pm \sqrt{529}}{10}\)

= \(\frac{-7 \pm 23}{10}=\frac{-7+23}{10}\)

= \(\frac{16}{10}\)[Taking sign]= \(\frac{8}{5}\)= \(1 \frac{3}{5}\)

Hence the speed of the stream is \(1 \frac{3}{5}\) km/hr

6. A superfast train runs having the speed 15 km/hr more than that of an express train. Leaving same station the superfast train reached at a station of 180 kms distance 1 hour before than the express train. Determine the speed of the superfast train in km/hr.

Solution:

Given

A superfast train runs having the speed 15 km/hr more than that of an express train. Leaving same station the superfast train reached at a station of 180 kms distance 1 hour before than the express train.

Let the speed of the superfast train was x km/hr.

∴ the speed of the express train was (x- 15) km/hr.

∴ to travel 180 km, time taken by the superfast train is \(\frac{180}{x}\) hr and by express train is \(\frac{180}{x-15}\)

As per questions, \(\frac{180}{x-15}\) – \(\frac{180}{x}\) = 1

or, \(180\left(\frac{1}{x-15}-\frac{1}{x}\right)=1\)

or, \(\frac{x-x+15}{(x-15) x}=\frac{1}{180}\)

or, \(\frac{15}{x^2-15 x}=\frac{1}{180}\)

or, x²-15x = 2700

or, x²-15x – 2700 = 0 ….(1)

Comparing (1) with ax²+bx+c=0, (a≠0) we get a=1,b = -15 and c = -2700

∴ by Sreedhar Acharya’s formula we get,

x= \(\frac{-(-15) \pm \sqrt{(-15)^2-4 \times 1 \times(-2700)}}{2 \times 1}\)

or, \(x=\frac{15 \pm \sqrt{225+10800}}{2}\)

or, \(x=\frac{15 \pm \sqrt{11025}}{2}\)

or, \(x=\frac{15 \pm 105}{2}\)

\(x =\frac{15+105}{2} \text { (taking }+ \text { sign) and } x=\frac{15-105}{2}(\text { taking }- \text { sign) }\)

= \(\frac{120}{2} \text { and } x=\frac{-90}{2}\)

=60 and x=-45

But the speed of a train cannot be negative, x = 60.

Hence the speed of the superfast train was 60 km/hr

Algebra Chapter 1 Quadratic Equation In One Variable Nature Of Roots

Let x² + 1 = 0 is a quadratic equation. From this we get, x² = – 1 or x = ± √-1. But √-1 is not a real number, i.e., the roots of this equation is not real number. It is imaginary.

Thus, the roots of a quadratic equation in one variable is not always real.

So, it is a matter of thinking that is it possible .to determine whether the roots of a quadratic equation in one variable are real or unreal or not, without finding the roots of the equation.

In reply we see that in Sreedhar Acharya’s formula if the quadratic equation be of the form ax² + bx + c = 0, then the roots will be

x \(=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)…..(1)

Now, in (1), all of a, b, and c are real and there is only one root-square \(\sqrt{b^2-4 a c}\) in this quantity.

Then if the value of (b² -4ac) be 0 or any other positive quantity, then we can easily do the square-root of that quantity and get a real number as a result.

But if the value of the quantity (b² – 4ac) be a negative number, i.e., if b² – 4ac < 0, then taking square root of the quantity, we do not get any real number.

In these cases, we get an unreal or complex number.

This (b²– 4ac) quantity is called the discriminant of the quadratic equation ax² + bx + c = 0 (a≠0)

Hence if the discriminant (b² – 4ac) of a quadratic equation be-

≥ 0, i.e., if b²– 4ac ≥ 0, then the roots of the given equation will be real;
< 0, i.e., if b²-4ac < 0, then the roots of the given equation will be unreal or imginary or complex.

Again, if 1. b²– 4ac = 0, then the roots of the given equation are real, equal, and rational.

Such as, the discriminant of the quadratic equation x² + 2x + 1 = 0 is equal to 2² – 4.1.1 =0.

∴ the roots of the equation x² + 2x + 1 = 0 are real, equal and rational.

2. if b²– 4ac > 0, then the roots of the quadratic equation in one variable ax² + bx + c = 0 will be real, unequal, and irrational.

3. The irrational roots of the equation ax²+ bx + c = 0 (a≠0) occurs in pair-wise, i.e., if one root of the equation be irrational, then the other root will also be irrational.

∴ if p +√q, q ∉I be a root of the equation, then the other root will be conjugate irrational, i.e. of the form p-√q , q∉l.

4. However, if b²– 4ac < 0, then the roots of the given equation will be imaginary, unequal, and irrational. The unreal roots also occur pair-wise, i.e., if one root of the given equation be (α+ iβ), then the other root will be also the conjugate imaginary, unequal, and irrational.

∴ if one root of the given quadratic equation be α+ iβ, then its other root will be of the form α- iβ

Here, (α- iβ) is called the complex conjugate of (α+ iβ) and vice versa.

From the above discussion we can say that without solving a given quadratic equation in one variable, only determining its discriminant and observing the value of it, we can say or determine the nature of the roots of the given quadratic equation.

To find the quadratic equation if its roots are given

Let α and β be two given roots of a quadratic equation. Then the equation will be x² – (α + β )x + αβ = 0.

For example, if two given roots of a quadratic equation be 2 and 3, then the equation will be x²-(2 + 3)x + 2 x 3 = 0 or, x² – 5x + 6 = 0

Relation between roots and coefficients of the quadratic equation ax²+bx+c = 0(a ≠ 0)

If α and β be two roots of the quadratic equation ax²+ bx + c = 0 (a≠0) then

1. α + β = –\(\frac{b}{a}\)

2. αβ= –\(\frac{c}{a}\) where a, b and c are the coefficients of x², x¹ and x° respectively.

For example, let a and (3 be the roots of the quadratic equation x² + 6x – 16 = 0, then α + β = – \(\frac{6}{1}\)

and αβ = – \(\frac{16}{1}\) or, αβ=-16 (Here, a = 1, b = 6 arid c =-16)

Proof of 1.

From Sreedhar Acharya’s formula we know that the roots of the quadratic equations are

\(\frac{-b+\sqrt{b^2-4 a c}}{2 a} \text { and } \frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

Now, let \(\alpha=\frac{-b+\sqrt{b^2}-4 a c}{2 a}\) and \(\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

∴ \(\alpha+\beta=\frac{-b+\sqrt{b^2-4 a c}}{2 a}+\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

= \(\frac{-b+\sqrt{b^2-4 a c}-b-\sqrt{b^2-4 a c}}{2 a}=\frac{-2 b}{2 a}=-\frac{b}{a}\)

∴ α + β = –\(\frac{b}{a}\)

Proof of 2.

\(\alpha \beta=\frac{-b+\sqrt{b^2-4 a c}}{2 a} \times \frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

= \(\frac{(-b)^2-\left(\sqrt{b^2-4 a c}\right)^2}{(2 a)^2}=\frac{b^2-\left(b^2-4 a c\right)}{4 a^2}\)

= \(\frac{b^2-b^2+4 a c}{4 a^2}=\frac{4 a c}{4 a^2}=\frac{c}{a}\)

∴\(\alpha \beta=\frac{c}{a}\)

Algebra Chapter 1 Quadratic Equation In One Variable Nature Of Roots Multiple Choice Questions

Example 1. The sum of the roots of the quadratic equation x²-6x+2 = 0 is

2
-2
6
-6

Solution: The sum of the roots of the equation

=\(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2} \)

= \(-\frac{-6}{1}=6\)

∴ 3. 6 is correct.

2. If the product of the roots of the quadratic equation x²– 3x + k- 10 be (- 2), then the value of k is

-2
– 8
8
12

Solution: The product of the roots of the equation x²-3x + k = 10 or, x² – 3x + k – 10 = 0

= \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

= \(\frac{k-10}{1}=k-10\)

As per question, k- 10 = – 2 or, k = – 2 + 10 = 8.

∴ 3. 8 is correct.

3. If the roots of the equation ax²+bx + c =0 (a ≠ 0) be equal, then

\(c=-\frac{b}{2 a}\)
\(c=\frac{b}{2 a}\)
\(c=\frac{-b^2}{4 a}\)
\(c=\frac{b^2}{4 a}\)

Solution: If the roots of the quadratic equation ax²+ bx + c= 0 (a≠ 0) be equal, then b²-4ac = 0 or b² = 4ac or \(c=\frac{b^2}{4 a}\)

∴ 4. \(c=\frac{b^2}{4 a}\)

4. If α and β be the roots of the q equation 3x² + 8x + 2 = 0 then the value of \(f\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) is

Solution: If the roots of the equation 3x² + 8x + 2 = 0 be α and β, then α+ β= –\(\frac{8}{3}\) and αβ = \(\frac{2}{3}\)

∴ \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-\frac{8}{3}}{\frac{2}{3}}=-4\)

∴ 3. \(c=\frac{-b^2}{4 a}\)

Example 2. Write true or false

1. If α and β be the roots of the equation x² – px + q = 0, then (α^-1 + β^-1) = \(\frac{p}{q}\)

Solution: True Since the given equation is x²– px + q = 0 and a and 3 are the roots of this equation.

∴α+ β = p and αβ = q.

Now, \(\begin{array}{r}
\alpha^{-1}+\beta^{-1}=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta} \\
=\frac{\alpha+\beta}{\alpha \beta}=\frac{p}{q}
\end{array}\)

Hence the statement is true.

2. If the roots of the equation px² +px + r = 0 (p≠0) be equal, then q = 2√pr

Solution: False

If the roots of the equation px² + qx + r = 0 [p≠0] be equal, then (- q)² – 4pr = 0 or, q² – 4pr = 0

or, q² = 4pr or, q – ±2√pr

Hence the given statement is false.

Example 3. Fill In the Blanks

1. The ratio of the sum and the product of the roots of the equation 7x²-12x+18 = 0 is _____

Solution: The sum of roots of the equation 7x²-12x+18 = 0 is

\(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

= \(-\frac{-12}{7}=\frac{12}{7}\)

and the product of the roots = \(\frac{\text { constant term }}{\text { coefficient of } x^2}=\frac{18}{7}\)

∴ Sum:Product = \(\frac{12}{7}\) : \(\frac{18}{7}\) = 12:18 =2:3

2. If the roots of the equation ax² + bx + c = 0 (a ≠0) be reciprocal to each other, then c = _____

Solution: Let a be one of the roots of the equation ax² +bx + c = 0 (a≠0).

∴ \(\frac{1}{\alpha}\) is another root of the equation

∴ \(\alpha+\frac{1}{\alpha}=-\frac{b}{a}\)……(1)

and \(\alpha \cdot \frac{1}{\alpha}=\frac{c}{a} or, 1=\frac{c}{a}\)

∴ c = a

3. If the roots of the equation ax² + bx + c = 0 (a≠ 0) be negatively reciprocal to each other, then a + c =_____

Solution: Let a be a root of the equation ax²+ bx+ c = 0.

∴ another root = – \(\frac{1}{\alpha}\)

∴ \(\alpha+\left(-\frac{1}{\alpha}\right)=-\frac{b}{a}\) and \(\alpha \times-\frac{1}{\alpha}=\frac{c}{a}\)

or, \(\alpha-\frac{1}{\alpha}=-\frac{b}{a} \text { and }-1=\frac{c}{a}\)

or, c = -a or c+a = 0

∴ a+c= 0

4. If the sum of the roots of the equationx²– x = k (2x – 1) be zero, then k = ______

Solution: Given that x² – x = k (2x-1)

or, x² – x = 2kx – k

or, x² – x – 2kx+ k = 0

or, x² – (2k + 1)x + k = 0 ….(1)

Now, the sum of the roots of the equation x² – (2k + 1)x + k = 0 is zero.

∴ \(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}=0\)

or, \(-\frac{-(2 k+1)}{1}=0\) or, 2 k+1=0

or, 2 k=-1 or, k=\(-\frac{1}{2}\)

∴ k=\(-\frac{1}{2}\)

Algebra Chapter 1 Quadratic Equation In One Variable Nature Of Roots Short Answer Type Questions

Example 1. The sum of the roots of a quadratic equation is 14 and the product is 24, then find the quadratic equation.

Solution: We know, the equation is x² -(sum of the roots) x + (product of the roots) = 0

∴ x²– 14x + 24 = 0 [v sum of the roots =14 and product of the roots = 24.]

Hence the required equation is x² – 14x + 24 = 0

Example 2. If the sum of the roots and product of the roots of the quadratic equation kx²+ 2x + 3k= 0 (k≠ 0) are equal, then find the value of k.

Solution: The sum of the roots of the equation –

kx²+ 2x + 3k= 0 is equal to \(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}=-\frac{2}{k}\)

and product of the roots = \(\frac{\text { constant term }}{\text { coefficient of } x^2}=\frac{3 k}{k}=3\)

As per question, – \(\frac{2}{k}\) = 3 or, 3k = – 2 or, k = – \(\frac{2}{k}\)

Hence k = – \(\frac{2}{k}\)

Example 3. If the roots of the equation x²-22x + 105 = 0 be α and β, then find the value of (α – β).

Solution: Let the root of the equation x²– 22x + 105 = 0 be α and β.

α + β = \(-\frac{22}{1}\) = 22 and αβ = \(\frac{105}{1}\) = 105.

Now, (α -β)² =(α + β)²-4αβ

= (22)² – 4 x 105 = 484 – 420 = 64.

∴ α – β= √64 = ± 8.

Example 4. If one root of the equations x² + bx + 12 = 0 and x² + bx + q = 0 be 2, then find the value of q.

Solution: One root of the equation x² + bx + 12 = 0 be 2,

∴ 2² + b.2 + 12 = 0 or, 4 + 2b +12 = 0 or, 2b +16 =0

or, 2 b= – 16 or, b = – \(\frac{16}{2}\) or, b = -8

Again, one root of the equation x² + bx + q = 0 is 2.

∴ 2² + b.2 + q = 0 or, 4 + 2b +q= 0 or,4+2 x -8+q=0

or, 4 – 16 + q = 0 or, – 12 + q = 0 or, q = 12.

Hence q = 12.

Example 5. If the roots of the equation ax²+bx+c = 0 be α and β, then find \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\)

Solution: Since the roots of the equation ax² + bx + c = 0 are α and β

∴ α + β = \(\frac{b}{a}\) and α β = \(\frac{c}{a}\)

Now,\(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\)

= \(\frac{\alpha^3+\beta^3}{\alpha \beta}\)

= \(\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}\)

= \(\frac{\left(-\frac{b}{a}\right)^3-3 \times \frac{c}{a} \times\left(-\frac{b}{a}\right)}{\frac{c}{a}}\)

=\(\frac{-\frac{b^3}{a^3}+\frac{3 b c}{a^2}}{\frac{c}{a}}\)

=\(\frac{\frac{-b^3+3 a b c}{a^3}}{\frac{c}{a}}\)

= \(\frac{-b^3+3 a b c}{a^3} \times \frac{a}{c}\)

= \(\frac{-b^3+3 a b c}{a^2 c}\)

∴ \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\)

∴ \(\frac{3 a b c-b^3}{a^2 c}\)

Example 6. If the product of the roots of the equation x² -3kx+2e^2Xogek -1 = 0 be 7, then find the value of k.

Solution: Given that x² – 3kx + 2e^2log_e^k -1 = 0

or, x² – 3kx + 2e^2log_e^k2 -1 = 0

or, x² – 3 kx + 2k² -1=0

∴ \(e^{\log _e k^2}=k^2\)….(1)

The product of the equation (1) \(\frac{2 k^2-1}{1}=2 k^2-1\)

As per question, 2k² -1=7

or, 2 k² = 7+1 or, 2 k² = 8 or, k² = 4 or, k ± 2.

But the value of k can never be negative, since log of any negative number is undefined.

∴ k≠ – 2, i.e., k = 2.

Hence the required value of k = 2

Example 7. If one root of a quadratic equation be (3 + √2), then find the equation.

Solution: One of the roots of the given equation is (3 + √2), which is an irrational number.

∴ The another root of the equation must be the conjugate irrational number of (3 + √2), i.e., the other root of the equation is (3-√2).

∴ The sum of the roots = 3 + √2+3-√2=6 and the product of the roots = (3 + √2) (3 – √2)

∴ (3)² -(√2)² =9-2-7

Hence the required equation is x² – 6x + 7 = 0.

Example 8. Find the roots of the quadratic equation |x|²-3|x| + 2 = 0

Solution: We know that | x | =

x, when x>0
0, when x = 0
-x when x<0 by the definition of absolute functions.

∴ from the given equation |x|²-3|x| + 2 = 0 we get x²-3x+ 2 = 0 when x>0

or, x²-x-2x + 2 = 0 or, x(x-1)-2(x-1) = 0

or, (x-1)(x-2) = 0

∴ either x-1 = 0 or, x-2 = 0

⇒ x = -1 or, x = -2

∴ x = ±1 and x = ±2

Hence the roots of the given equations are ±1 and x = ±2

Algebra Chapter 1 Quadratic Equation In One Variable Nature Of Roots Long Answer Type Questions

Example 1. If the roots of the quadratic equation (a²+ b²) x² – 2 (ac + bd) x + (c² + d²) = 0 be equal, then prove that \(\frac{a}{b}\) = \(\frac{c}{d}\)

Solution: The roots of the given equation are equal.

The discriminant of the equation must be zero,

i.e., {- 2(ac + bd)}² – 4(a² + b²)(c² + d²) = 0

or, 4 (ac + bd)² – 4 (a²c² + b²c² + a²d² + b²d²) = 0

or, 4 (a²c² + 2.ac.bd + b²d²) – 4 (a²c² + b²c² + a²d² + b²d²) = 0

or, 4 a²c² + 8abcd + 4b²d² – 4 a²c² – 4b²c² – 4a²d² – 4b²d²) = 0

or, 8abcd – 4a²d² – 4b²c² = 0

or, – (4a²d² – 8abcd + 4b²c²) = 0

or, 4 (a²d² – 2abcd + b²c²) = 0

or, (ad)² – 2.ad.bc + (bc)² = 0

or, (ad – bc)² = 0 or, ad – bc = 0

or, ad = bc or, \(\frac{a}{b}\)= \(\frac{c}{d}\)

Hence proved.

Example 2. Prove that the quadratic equation 2(a² + b²) x² + 2 (a + b) x + 1 = 0 will have no real root if a≠b.

Solution: The given equation will have no real root if its discriminant < 0, i.e., if

{2 (a + b)}² -4 x 2(a²+b²) x 1<0.

⇒ 4 (a + b)² – 8 (a² + b²) < 0

⇒ 4 (a² + 2ab + b²) – 8a² – 8b² < 0

⇒ 4a² + 8ab + 4b² – 8a² – 8b² < 0

⇒ – 4 a² + 8 ab – 4b² <0 ⇒ – 4 (a² – 2ab + b²) < 0

⇒ – (a² -2ab + b²) < 0 ⇒ – (a – b)² < 0

But (a – b)² is either 0 or any positive number. So, – (a – b)² < 0 is true if a – b = 0 or, a = b.

∴ if a = b, then the given equation will have real roots.

Example 3. If α and β be the roots of the equation 5x² + 2x – 3 = 0, then find the values of

α² + β²
α³ + β³
\(\frac{1}{\alpha}+\frac{1}{\beta}\)
\(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\)

Solution: 1. Since α and β are the roots of the equation 5x² + 2x – 3 = 0 then

\(\alpha+\beta=-\frac{2}{5} \text { and } \alpha \beta=\frac{-3}{5}\) [as per formula]

Now, α² + β² = (α + β)²– 2αβ

\(\begin{aligned}
& =\left(-\frac{2}{5}\right)^2-2 \times \frac{-3}{5} \\
& =\frac{4}{25}+\frac{6}{5}=\frac{4+30}{25}=\frac{34}{25}
\end{aligned}\)

Here, α² + β² = \(\frac{34}{25}\)

2. α³ + β³ = (α + β)³-3αβ(α + β)³

\(\begin{aligned}
& =\left(-\frac{2}{5}\right)^3-3 \times \frac{-3}{5} \times \frac{-2}{5} \\
& =\frac{-8}{125}-\frac{18}{25}=\frac{-8-90}{125}=\frac{-98}{125}
\end{aligned}\)

Hence α³ + β³ = – \(\frac{98}{125}\)

3. \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{\alpha+\beta}{\alpha \beta}\)

=\(\frac{-\frac{2}{5}}{-\frac{3}{5}}=\frac{2}{3}\)

Hence \(\frac{1}{\alpha}+\frac{1}{\beta}\) = \(\frac{2}{3}\)

4. \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} =\frac{\alpha^3+\beta^3}{\alpha \beta}\)

=\(\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}\)

= \(\frac{\left(-\frac{2}{5}\right)^3-3 \times \frac{-3}{5} \times \frac{-2}{5}}{-\frac{3}{5}}\)

\(\left[because \alpha+\beta=\frac{-2}{5} \text { and } \alpha \beta=\frac{-3}{5}\right]\)

= \(\frac{-\frac{8}{125}-\frac{18}{25}}{-\frac{3}{5}}\)

=\(\frac{\frac{-8-90}{125}}{-\frac{3}{5}}=\frac{-\frac{98}{125}}{-\frac{3}{5}}\)

=\(-\frac{98}{125} \times-\frac{5}{3}=\frac{98}{75}\)

Hence \(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\) = \(\frac{98}{75}\)

Example 4. If one root of the equation ax² + bx + c = 0 be double the another root of the equation, then prove that 2b² = 9ac.

Solution: Let one of the roots of the given equation be α.

As per question, the other root is 2α.

So, \(\alpha+2 \alpha=-\frac{b}{a}\)

or, \(3 \alpha=-\frac{b}{a}\)

or, \(\alpha=-\frac{b}{3 a}\)…(1)

Again \( \alpha \times 2 \alpha=\frac{c}{a}\)

or, \( 2 \alpha^2=\frac{c}{a}\)

or, \(2 \times\left(-\frac{b}{3 a}\right)^2=\frac{c}{a}\)

or, \(2 \times \frac{b^2}{9 a^2}=\frac{c}{a}\)

or, \(\frac{2 b^2}{9 a}=c\)

or, \(2 b^2=9 a c\)

Hence \(2 b^2=9 a c\)(Proved)

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