# R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations

R S Aggarwal Solutions for Class 10 Chapter 4 Quadratic Equations is a set of important questions which helps students to master quadratic equations. These solutions help students in preparing for the exams in a better way. Download RS Aggarwal Class 10 Maths Solutions for Chapter 4. Basically, this study material is aimed at mastering the concepts and acquiring comprehensive knowledge about the varied types of questions asked in CBSE Class 10 Maths exams.

### Access Answers to Maths R S Aggarwal Chapter 4 Quadratic Equations

Get detailed solutions for all the questions listed under the below exercises:

Exercise 4A Solutions: 30 Questions (Short Answers)

Exercise 4B Solutions: 16 Questions (Short Answers)

Exercise 4C Solutions: 15 Questions (Short Answers)

Exercise 4D Solutions: 20 Questions (Long Answers)

## Exercise 4A Page No: 175

Question 1:

Solution:

AÂ quadratic equationÂ is anÂ equationÂ of the second degree.

(i) x2 â€“ x + 3 = 0

Highest Degree: 2

(ii) 2x2 +Â 5/2x – âˆš3 = 0

Highest Degree: 2

(iii) âˆš2x2+7x+5âˆš2=0

Highest Degree: 2

(iv) 1/3x2 + 1/5x â€“ 2 = 0

Above equation can be simplify as: 5x2 + 3x â€“ 2 = 0

Highest Degree: 2

(v) x2 – 3x – âˆšx + 4 = 0

Equation has a fractional power.

(vi) x â€“ 6/x = 3

Simply as x2 – 3x – 6 = 0

Degree : 2

(vii) x + 2/x = x2

Simplify above equation:

x3 – x2 – 2 = 0

Degree: 3

(viii) x^2 â€“ 1/x^2 = 5

Simplify above equation

x4 Â  – 1 = 5 x2

or x4-5x2-1 =0

Degree: 4

(ix) (x + 2)^3 = x^3 â€“ 8

x^3 + 8 + 6x^2 + 12x = x^3 â€“ 8

-6x^2 + 12x + 16 = 0

Degree = 2

(x) (2x + 3)(3x + 2) = 6(x â€“ 1)(x â€“ 2)

Simplify above equation:

6x^2 + 4x + 9x + 6 = 6x^2 â€“ 12x â€“ 6x + 12

31x â€“ 6 = 0

Degree : 1

(xi) (x + 1/x)^2 = 2)x + 1/x) + 3

Simplify above equation:

(x^4 + 2x^2 + 1) / x^2 = (2x^2 + 2) / x + 3

(x^4 + 2x^2 + 1)x = x^2(2x^2 + 2) + 3

Answer: (i), (ii), (iii), (iv), (vi) and (ix) are only quadratic equations.

Question 2:

Solution:

Simplify given equation:

3xÂ² + 2x â€“ 1 = 3xÂ² + 3x â€“ x â€“ 1

= 3x (x + 1) â€“ 1 (x + 1)

= (x + 1) (3x â€“ 1)

To find roots, put 3xÂ² + 2x â€“ 1 = 0

Either, x + 1 = 0 or 3x â€“ 1 =0

x = -1 or x = 1/3

Therefore, (-1) and 1/3 are the required roots.

Question 3:

Solution:

(i) x = 1 is a solution of x^2+kx+3=0, which means it must satisfy the equation.

(1)^2 + k(1) + 3 = 0

k = -4

Hence the required value of k = -4

Find other root:

We have equation, x^2 – 4x + 3 = 0

x^2 – x – 3x + 3 = 0

x(x – 1) – 3(x – 1) = 0

(x – 1)(x – 3) = 0

either x – 1 = 0 or x – 3 = 3

x = 1 or x = 3

Other root is 3.

(ii) given equation is axÂ² + bx â€“ 6 = 0

As Â¾ is its root, then must satisfy the equation

a(3/4)Â² + b(3/4) â€“ 6 = 0

9a +12b â€“ 96 = 0 â€¦.(1)

Again, x = -2 is its root

a(-2)Â² + b(-2) â€“ 6 = 0

4a â€“ 2b â€“ 6 = 0 â€¦(2)

Solving (1) and (2), we get

a = 4 and b = 5

Question 4:Â

Question 5:

Solution:

(2x â€“ 3)(3x + 1) = 0

Either 2x â€“ 3 = 0 or 3x + 1 = 0

x = 3/2 or x = -1/3

Question 6:Â

Solution:

4x2 + 5x = 0

Or x (4x + 5) = 0

Either x = 0 or 4x + 5 = 0, then

x = -5/4 or 0

Question 7:

Solution:

3xÂ² â€“ 243 = 0

or xÂ² â€“ 81 = 0

(x)Â² â€“ (9)Â² = 0

(x + 9) (x â€“ 9) = 0

Either, x + 9 = 0 or x â€“ 9 = 0

x = -9 or 9

Question 8:Â

Solution:

2x2 + x – 6 = 0

2x2 + 4x – 3x – 6 = 0

2x (x + 2) – 3(x + 2) = 0

(x + 2)(2x – 3)= 0

Either x + 2 = 0 or 2x – 3 = 0

x = -2 or 3/2

Question 9:Â

Solution:

x^2 + 6x + 5 = 0

x^2 + x + 5x + 5 = 0

x(x + 1) + 5(x + 1) = 0

(x + 5)(x + 1) = 0

either x +5 = 0 or x + 1 = 0

x = -5 or -1

Question 10:

Solution:

9x^2 â€“ 3x â€“ 2 = 0

9x^2 â€“ 6x + 3x â€“ 2 = 0

3x(3x -2) + (3x â€“ 2) = 0

(3x +1)(3x â€“ 2) = 0

either (3x +1) = 0 or (3x â€“ 2) = 0

x = -1/3 or 2/3

Question 11:Â

Solution:

xÂ² + 12x + 35 = 0

xÂ² + 7x + 5x + 35 = 0

x(x + 7) + 5((x + 7) = 0

(x + 5)(x + 7) = 0

either (x + 5) = 0 or (x + 7) = 0

x = -5 or -7

Question 12:Â

Solution:

x^2 – 18x + 77 = 0

x^2 – 7x – 11x + 77 = 0

x (x – 7) – 11(x – 7) = 0

(x – 11)(x – 7) = 0

either (x – 11) = 0 or (x – 7) = 0

x = 11 or 7

Question 13:Â

Solution:

6x^2 + 11x + 3 = 0

6x^2 + 2x + 9x + 3 = 0

2x(3x + 1) + 3(3x + 1) = 0

(2x + 3)(3x + 1) = 0

either (2x + 3) = o or (3x + 1) = 0

x = -1/3 or -3/2

Question 14:

Solution:

Question 15:

Solution:

Question 16:

Solution:

4xÂ² â€“ 9x = 100

4xÂ² â€“ 9x â€“ 100 = 0

Question 17:

Solution:

15x2 ï€­ 28 ï€½ x

15x2 ï€­ x ï€­ 28 ï€½ 0

15x2 ï€­ ï€¨21x ï€­ 20xï€©ï€­ 28 ï€½ 0

15x2 ï€­ 21x ï€« 20x ï€­ 28 ï€½ 0

3x ï€« 4 ï€½ 0 or 5x ï€­ 7 ï€½ 0

x = -4/3 or 7/5

Question 18:

Solution:

4 ï€­11x ï€½ 3x2

3x2 ï€«11x ï€­ 4 ï€½ 0

3x2 ï€«12x ï€­ x ï€­ 4 ï€½ 0

Either x ï€« 4 ï€½ 0 or 3x ï€­1 ï€½ 0

x = -4 or 1/3

Question 19:Â

Solution:

48x2 ï€­13x ï€­1 ï€½ 0

48x2 ï€­ï€¨16x ï€­ 3xï€©ï€­1 ï€½ 0

48x2 ï€­16x ï€« 3x ï€­1 ï€½ 0

Either 16x ï€«1 ï€½ 0 or 3x ï€­1 ï€½ 0

x = -1/16 or 1/3

Question 20:

Solution:

Question 21;

Solution:

âˆš3 xÂ² + 10x + 7âˆš3 = 0

âˆš3 xÂ² + 10x + 7âˆš3 = 0

âˆš3 xÂ² + 3x + 7x + 7âˆš3 = 0

âˆš3 x(x + âˆš3) + 7( x + âˆš3) = 0

(x + âˆš3)(âˆš3 x + 7) = 0

either âˆš3 x + 7 = 0 or x + âˆš3 = 0

x = -âˆš3 or -7/âˆš3

Question 22:

Solution:

âˆš3 xÂ² + 11x + 6âˆš3 = 0

âˆš3 xÂ² + 9x + 2x + 6âˆš3 = 0

âˆš3 x (x + 3âˆš3) + 2(x + 3âˆš3) = 0

(âˆš3x + 2)(x + 3âˆš3) = 0

either (âˆš3x + 2) = 0 or (x + 3âˆš3) = 0

x = -3âˆš3 or -2âˆš3/3

Question 23:

Solution:

3âˆš7 xÂ² + 4x – âˆš7 = 0

3âˆš7 xÂ² + 4x – âˆš7 = 0

3âˆš7 xÂ² -3x + 7x – âˆš7 = 0

3x(âˆš7x – 1) + âˆš7(âˆš7x – 1) = 0

(3x + âˆš7)(âˆš7x – 1) = 0

either (3x + âˆš7) = 0 or (âˆš7x – 1) = 0

x = -âˆš7/3 or 1/âˆš7

Question 24:

Solution:

âˆš7 xÂ² â€“ 6x â€“ 13âˆš7 = 0

Question 25:

Solution:

Question 26:Â

Solution:

3xÂ² â€“ 2âˆš6x + 2 = 0

Question 27:Â

Solution:

âˆš3 xÂ² â€“ 2âˆš2 x â€“ 2âˆš3 = 0

âˆš3 xÂ² â€“ 3âˆš2 x + âˆš2 xâ€“ 2âˆš3 = 0

âˆš3x (x – âˆš6) + âˆš2(x – âˆš6) = 0

(âˆš3x+ âˆš2)(x – âˆš6) = 0

either (âˆš3x+ âˆš2) = 0 or (x – âˆš6) = 0

x = âˆš6 or -âˆš2/âˆš3

Question 28:Â

Solution:

Question 29:

Solution:

xÂ² â€“ (âˆš3 + 1) x + âˆš3 = 0

xÂ² â€“ (âˆš3 + 1) x + âˆš3 = 0

xÂ² â€“ âˆš3x – x + âˆš3 = 0

x(x – âˆš3) – (x – âˆš3) = 0

(x – 1)(x – âˆš3) = 0

either (x – 1) = 0 or (x – âˆš3) = 0

x = 1 or âˆš3

Question 30:

Solution:

xÂ² + 3âˆš3 x â€“ 30 = 0

xÂ² + 5âˆš3 x – 2 âˆš3 â€“ 30 = 0

x(x + 5âˆš3) – 2âˆš3 (x + 5âˆš3) = 0

(x – 2âˆš3)(x + 5âˆš3) = 0

either (x – 2âˆš3) = 0 or (x + 5âˆš3) = 0

x = -5âˆš3 or 2 âˆš3

## Exercise 4B Page No: 185

Solve each of the following equations by using the method of completing the square:

Question 1:

Solution:

xÂ² â€“ 6x + 3 = 0

x^2 – 6x + 3 = 0

x^2 – 6x = – 3

x^2 – 2(x)3 + 3^2 = -3 + 3^2

(x – 3)^2 = – 3 + 9 = 6

Using algebraic identity: a^2 – 2ab + b^2 = (a – b)^2

Question 2:

Solution:

x^2 – 4x + 1 = 0

x^2 – 4x = – 1

x^2 – 2(x)(2) + 2^2 = – 1 + 2^2

(x – 2)^2 = 3

Using algebraic identity: a^2 – 2ab + b^2 = (a – b)^2

Question 3:

Solution:

x^2 + 8x â€“ 2 = 0

x^2 + 8x = 2

x^2 – 2.x.4 + 42 = 2 + 42

(x + 4)^2 = 18

Using algebraic identity: a^2 – 2ab + b^2 = (a – b)^2

Question 4:

Solution:

Question 5:

Solution:

2x^2 + 5x – 3 = 0

4x^2 + 10x – 6 = 0

(multiplying both sides by 2)

4x^2 + 10x = 6

(2x + 5/2)^2 = 6 + 25/4 = 49/4

Using algebraic identity: a^2 – 2ab + b^2 = (a – b)^2

Taking square root,

2x + 5/2 = 7/2 or 2x + 5/2 = -7/2

x = 1/2 or -3

Question 6:

Solution:

3x^2 – x – 2 = 0

9x^2 – 3x – 6 = 0

(multiplying both sides by 3)

9x^2 – 3x = 6

Adding (1/2)^2 on both the sides.

3x – 1/2 = 5/2 or 3x – 1/2 = -5/2

x = 1 or x = -2/3

Question 7:

Solution:

8x^2 – 14x – 15 = 0

16x^2 – 28x – 30 = 0

(multiplying both sides by 2)

Adding (7/2)^2 on both the sides

4x – 7/2 = 13/2 or 4x – 7/2 = -13/2

x = 5/2 or x = -3/4

Question 8:

Solution:

7x^2 + 3x – 4 = 0

49x^2 + 21x – 28 = 0

(multiplying both sides by 7)

Adding (3/2)^2 on both the sides,

7x + 3/2 = 11/2 or 7x + 3/2 = -11/2

x = -1 or x = 4/7

Question 9:

Solution:

3x^2 – 2x – 1 = 0

9x^2 – 6x = 3

(multiplying both sides by 3)

Adding (1)^2 on both the sides

(3x)^2 – 2.3x.1 + (1)^2 = 3 + (1)^2

(3x – 1)^2 = 2^2

3x – 1 = 2 or 3x – 1 = -2

x = -1 or x = -1/3

Question 10:

Solution:

5x^2 – 6x – 2 = 0

25x^2 – 30x – 10 = 0

(multiplying both sides by 5)

25x^2 – 30x = 10

Question 11:

Solution:

Question 12:

Solution:

Question 13:

Solution:

Question 14:

Solution:

âˆš2 xÂ² â€“ 3x â€“ 2âˆš2 = 0

Dividing each side by âˆš2

Question 15:

Solution:

âˆš3 xÂ² + 10x â€“ 7âˆš3 = 0

Dividing each side by âˆš3

Question 16:

Solution:

## Exercise 4C Page No: 191

Find the discriminant of each of the following equations:

Question 1:

Solution:

(i) 2xÂ² â€“ 7x + 6 = 0

Compare given equation with the general form of quadratic equation, which is

ax^2 + bx + c = 0

Here, a = 2, b = -7 and c = 6

Discriminant formula: D = b^2 – 4ac

(-7)^2 – 4 x 2 x 6

= 1

(ii) 3xÂ² â€“ 2x + 8 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 3, b = â€“ 2, c = 8

Discriminant formula: D = b^2 – 4ac

= (â€“ 2)^2 â€“ 4.3.8

= 4 â€“ 96

= â€“ 92

(iii) 2xÂ² â€“ 5âˆš2x + 4 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 2, b = â€“ 5âˆš2, c = 4

Discriminant formula: D = b^2 – 4ac

= (â€“ 5âˆš2)^2 â€“ 4.2.4

= 50 – 32

= 18

(iv) âˆš3 xÂ² + 2âˆš2 x â€“ 2âˆš3 =0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = âˆš3, b = 2âˆš2, c = â€“ 2âˆš3

Discriminant formula: D = b^2 – 4ac

= (2âˆš2)^2 â€“ 4(âˆš3)(â€“ 2âˆš3)

= 32

(v) (x â€“ 1) (2x â€“ 1) = 0

2x^2 â€“ 3x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 2, b = -3, c = â€“1

Discriminant formula: D = b^2 – 4ac

= (-3)^2 â€“ 4x2x1

= 1

(vi) 1 â€“ x = 2xÂ²

1 â€“ x = 2xÂ²

2x^2 + x â€“ 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 2, b = 1, c = â€“1

Discriminant formula: D = b^2 – 4ac

= (1)^2 â€“ 4x2x-1

= 9

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

Question 2:

Solution:

xÂ² â€“ 4x â€“ 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 1, b = -4, c = â€“1

Find Discriminant:

D = b^2 – 4ac

= (-4)^2 â€“ 4x1x-1

= 20 > 0

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Therefore, x = 2 + âˆš5 and x = 2 – âˆš5

Question 3:

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = -6, c = 4

Find Discriminant:

D = b^2 – 4ac

= (-6)^2 â€“ 4.1.4

= 36 – 16

= 20 > 0

Roots of equation are real.

Find the Roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

x = 3 + âˆš5 or x = 3 â€“ âˆš5

Question 4:

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = 1, c = -4

Find Discriminant:

D = b^2 – 4ac

= (1)^2 â€“ 4.2.-4

= 1 + 32

= 33 > 0

Roots of equation are real.

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Question 5:

Solution:

25xÂ² + 30x + 7 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 25, b = 30, c = 7

Find Discriminant:

D = b^2 – 4ac

= (30)^2 â€“ 4.25.7

= 900 â€“ 700

= 200 > 0

Roots of equation are real.

Find the Roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Question 6:

Solution:

16xÂ² – 24x â€“ 1 =0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 16 , b = -24, c = -1

Find Discriminant:

D = b^2 – 4ac

= (-24)^2 â€“ 4.16.-1

= 576 + 64

= 640 > 0

Roots of equation are real.

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Question 7:

Solution:

15xÂ² -x â€“ 28 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 15, b =-1 , c = -28

Find Discriminant:

D = b^2 – 4ac

= (-1)^2 â€“ 4.15.(-28)

= 1 + 1680

= 1681 > 0

Roots of equation are real.

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

X = 7/5 or x = -4/3

Question 8:

Solution:

2xÂ² â€“ 2âˆš2 x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = â€“ 2âˆš2, c = 1

Find Discriminant:

D = b^2 – 4ac

= (â€“ 2âˆš2)^2 â€“ 4.2.1

= 8 – 8

= 0

Equation has equal root.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

x = 1/âˆš2 or x = 1/âˆš2

Roots of equation are real.

Question 9:

Solution:

âˆš2 xÂ² + 7x + 5âˆš2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = âˆš2 , b = 7 , c = 5âˆš2

Find Discriminant:

D = b^2 – 4ac

= (7)^2 â€“ 4. âˆš2 . 5âˆš2

= 49 – 40

= 9> 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Roots are:

x = – âˆš2 or x = -5/âˆš2

Question 10:

Solution:

âˆš3 xÂ² + 10x â€“ 8âˆš3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = âˆš3, b = 10 , c = â€“ 8âˆš3

Find Discriminant:

D = b^2 – 4ac

= (10)^2 â€“ 4. âˆš3 . â€“ 8âˆš3

= 100 + 96

= 196 > 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Roots are:

x = 2âˆš3/3 or x = -4âˆš3

Question 11:

Solution:

âˆš3 xÂ² â€“ 2âˆš2 x â€“ 2âˆš3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = âˆš3 , b = â€“ 2âˆš2, c = â€“ 2âˆš3

Find Discriminant:

D = b^2 – 4ac

= (â€“ 2âˆš2)^2 â€“ 4. âˆš3 . â€“ 2âˆš3

= 8 + 24

= 32 > 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

x = âˆš6 or x = -âˆš2/âˆš3

Question 12:

Solution:

2xÂ² + 6âˆš3 x â€“ 60 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = 6âˆš3, c = â€“ 60

Find Discriminant:

D = b^2 – 4ac

= (6âˆš3)^2 â€“ 4.2. â€“ 60

= 108 + 480

= 588 > 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Question 13:

Solution:

4âˆš3 xÂ² + 5x â€“ 2âˆš3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 4âˆš3 , b = 5, c = â€“ 2âˆš3

Find Discriminant:

D = b^2 – 4ac

= (5)^2 â€“ 4. 4âˆš3 . â€“ 2âˆš3

= 25 + 96

= 121 > 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Roots are:

x = âˆš3/4 or x = -2/âˆš3

Question 14:

Solution:

3xÂ² â€“ 2âˆš6 x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 3, b = â€“ 2âˆš6 , c = 2

Find Discriminant:

D = b^2 – 4ac

= (â€“ 2âˆš6)^2 â€“ 4. 3 . 2

= 24 – 24

= 0

Roots of equation are equal.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Roots are:

x = âˆš2/âˆš3 and x = x = âˆš2/âˆš3

Question 15:

Solution:

2âˆš3 xÂ² â€“ 5x + âˆš3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2âˆš3, b = â€“ 5 , c = âˆš3

Find Discriminant:

D = b^2 – 4ac

= (â€“ 5)^2 â€“ 4. 2âˆš3 . âˆš3

= 25 – 24

= 1 > 0

Roots of equation are real.

Find roots:

$\large x = \frac{-b\pm \sqrt{D}}{2a}$

Roots are:

x = âˆš3/2 and x = 1/âˆš3

## Exercise 4D Page No: 199

Question 1:

Solution:

(i)2xÂ² â€“ 8x + 5 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = â€“ 8, c = 5

Using Discriminant Formula:

D = b^2 â€“ 4ac

= (â€“ 8)^2 â€“ 4.2.5

= 64 â€“ 40

= 24 > 0

Hence the roots of equation are real and unequal.

(ii) 3xÂ² â€“ 2âˆš6 x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 3, b = â€“ 2âˆš6, c = 2

Using Discriminant Formula:

D = b^2 â€“ 4ac

= (â€“ 2âˆš6)^2 â€“ 4.3.2

=24 – 24

= 0

Roots of equation are real and equal.

(iii) 5xÂ² â€“ 4x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 5, b = â€“ 4, c = 1

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 4)^2 â€“ 4.5.1

= 16 â€“ 20

= â€“ 4 < 0

Equation has no real roots.

(iv) 5x (x â€“ 2) + 6 = 0

5x^2 â€“ 10x + 6 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 5, b = â€“ 10, c = 6

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 10)2 â€“ 4.5.6

= 100 â€“ 120

= â€“ 20 < 0

Equation has no real roots.

(v) 12xÂ² â€“ 4âˆš15 x + 5 = 0

12xÂ² â€“ 4âˆš15 x + 5 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 12, b = â€“ 4âˆš15, c = 5

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 4âˆš15)^2 â€“ 4.12.5

= 240 – 240

= 0

Equation has real and equal roots.

(vi) xÂ² â€“ x + 2 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = â€“ 1, c = 2

Discriminant:

D = b^2 â€“ 4ac

= (â€“ 1)^2 â€“ 4.1.2

= 1 â€“ 8

= â€“ 7 < 0

Equation has no real roots.

Question 2:

Solution:

2(aÂ² + bÂ²) xÂ² + 2 (a + b) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2 (a^2 + b^2), b = 2(a + b), c = 1

Discriminant:

D = b^2 â€“ 4ac

=[2(a + b)]^2 â€“ 4. 2 (a^2 + b^2).1

= 4a^2 + 4b^2 + 8ab â€“ 8a^2 â€“ 8b^2

= â€“ 4a^2 â€“ 4b^2 + 8ab

= â€“ 4(a^2 + b^2 â€“ 2ab)

= â€“ 4(a â€“ b)^2 < 0

Hence the equation has no real roots.

Question 3:

Solution:

xÂ² + px â€“ qÂ² = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = p, c = -qÂ²

Using discriminant formula:

D = bÂ² â€“ 4ac

= (p)Â² â€“ 4 x 1 x (-qÂ²)

= pÂ² + 4 qÂ² > 0

Hence roots are real for all real values of p and q.

Question 4:

Solution:

3xÂ² + 2kx + 27 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 3, b = 2k, c = 27

Find discriminant:

D = bÂ² â€“ 4ac

= (2k)Â² â€“ 4 x 3 x 27

= (2k)Â² â€“ 324 >0

Roots are real and equal

Find the value of k:

(2k)Â² â€“ 324 = 0

(2k)Â² â€“ (18)Â² = 0

(k)Â² â€“ (9)Â² = 0

(k + 9) (k â€“ 9) = 0

Either k + 9 = 0 or k â€“ 9 = 0

k = -9 or k = 9

Hence, the value of k is k = 9 or -9

Question 5:

Solution:

kx (x â€“ 2âˆš5) x + 10 = 0

kxÂ² â€“ 2âˆš5 kx + 10 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = k, b = -2âˆš5 k, c = 10

Find discriminant:

D = bÂ² â€“ 4ac

= (-2 k)Â² â€“ 4 x k x 10 = 20kÂ² â€“ 40k

Since roots are real and equal (given), put D = 0

20kÂ² â€“ 40k = 0

kÂ² â€“ 2k = 0

k (k â€“ 2) = 0

Either, k = 0 or k â€“ 2 = 0

Hence k = 0 or k = 2

Question 6:

Solution:

4xÂ² + px + 3 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 4, b = p, c = 3

Find discriminant:

D = bÂ² â€“ 4ac

= pÂ² â€“ 4 x 4 x 3

= pÂ²- 48

Since roots are real and equal (given)

Put D = 0

pÂ² â€“ 48 = 0

pÂ² = 48 = (Â±4âˆš3)Â²

p = Â± 4âˆš3

Hence p= 4âˆš3 or p = -4âˆš3

Question 7:

Solution:

9xÂ² â€“ 3kx + k = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here a = 9, b = -3k, c = k

Find discriminant:

D = bÂ² â€“ 4ac

= (-3k)Â² â€“ 4 x 9 x k

= 9kÂ² â€“ 36k

Since roots are real and equal (given)

Put D = 0

9kÂ² â€“ 36k = 0

9k (k â€“ 4) = 0

Either, k = 0 or k â€“ 4 = 0

As, value of k is non-zero:

So, k = 4

Question 8:

Solution:

(i) (3k + 1) xÂ² + 2(k + 1) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (3k + 1), b = 2(k + 1), c = 1

Find discriminant:

D = bÂ² â€“ 4 ac

= (2(k + 1))Â² â€“ 4(3k + 1) x 1

= 4kÂ² + 4 + 8k â€“ 12k â€“ 4

= 4k (k â€“ 1)

Since roots are real and equal (given)

Put D = 0

4k (k â€“ 1) = 0

Either, k = 0 or k â€“ 1 = 0

k = 0, k = 1

(ii) xÂ² + k(2x + k â€“ 1) + 2 = 0

Simplify above equation:

xÂ² + 2kx + (kÂ² â€“ k + 2) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

Here, a = 1, b = 2k, c = (kÂ² â€“ k + 2)

Find Discriminant:

D = bÂ² â€“ 4ac

= (2k)Â² â€“ 4 x 1 x (kÂ² â€“ k + 2)

= 4kÂ² â€“ 4kÂ² + 4k â€“ 8

= 4k â€“ 8

Since roots are real and equal (given)

Put D = 0

4k â€“ 8 = 0

k = 2

Hence, the value of k is 2.

Question 9:

Solution:

(2p + 1) xÂ² â€“ (7p + 2) x + (7p â€“ 3) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (2p + 1), b = â€“ (7p + 2) and c = (7p â€“ 3)

Discriminant:

D = b^2 â€“ 4ac

= (â€“ (7p + 2))^2 â€“ 4.(2p + 1).(7p â€“ 3)

= (49p^2 + 28p + 4) â€“ 4(14p^2 + p â€“ 3)

= 49p^2 + 28p + 4 â€“ 56p^2 â€“ 4p + 12

= – 7p2 + 24p + 16

Since roots are real and equal (given)

Put D = 0

7p^2 â€“ 24p â€“ 16 = 0

7p^2 â€“ 28p + 4p â€“ 16 = 0

7p(p â€“ 4) + 4(p â€“ 4) = 0

(7p + 4)(p â€“ 4) = 0

Either (7p + 4) = 0 or (p â€“ 4) = 0

p = -4/7 or p = 4

Question 10:

Solution:

(p + 1) xÂ² â€“ 6(p + 1) x + 3(p + 9) = 0, p â‰  -1

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (p + 1), b = â€“ 6(p + 1) and c = 3(p + 9)

Discriminant:

D = b^2 â€“ 4ac

= (-6(p + 1))^2 â€“ 4.(p + 1).3(p + 9)

= 36(p + 1)(p + 1) â€“ 12(p + 1)(p + 9)

= 12(p + 1) (3p + 3 â€“ p â€“ 9)

= 12(p + 1)(2p â€“ 6)

Since roots are real and equal (given)

Put D = 0

12(p + 1)(2p â€“ 6) = 0

either (p + 1) = 0 or (2p â€“ 6) = 0

p = -1 or p = 3

Question 11:

Solution:

Given: -5 is a root of the quadratic equation 2xÂ² + px â€“ 15 = 0

Substitute the value of x = -5

2(-5)Â² + p(-5) â€“ 15 = 0

50 â€“ 5p â€“ 15 = 0

35 â€“ 5p = 0

p = 7

Again,

In quadratic equation p(xÂ² + x) + k = 0

7 (xÂ² + x) + k = 0 (put value of p = 7)

7xÂ² + 7x + k = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 7, b = 7, c = k

Find Discriminant:

D = bÂ² â€“ 4ac

= (7)Â² â€“ 4 x 7 x k

= 49 â€“ 28k

Since roots are real and equal, put D = 0

49 â€“ 28k = 0

28k = 49

k = 7 / 4

The value of k is 7/4

Question 12:

Solution:

Given: 3 is a root of equation xÂ² â€“ x + k = 0

Substitute the value of x = 3

(3)Â² â€“ (3) + k = 0

9 â€“ 3 + k = 0

k = -6

Now, xÂ² + k (2x + k + 2) + p = 0

xÂ² + (-6)(2x â€“ 6 + 2) + p = 0

xÂ² â€“ 12x + 36 â€“ 12 + p = 0

xÂ² â€“ 12x + (24 + p) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = -12, c = 24 + p

Find Discriminant:

D = bÂ² â€“ 4ac

= (-12)Â² â€“ 4 x 1 x (24 + p)

= 144 â€“ 96 â€“ 4p = 48 â€“ 4p

Since roots are real and equal, put D = 0

48 â€“ 4p = 0

4p = 48

p = 12

The value of p is 12.

Question 13:

Solution:

Given: -4 is a root of the equation xÂ² + 2x + 4p = 0

Substitute the value of x = -4

(-4)Â² + 2(-4) + 4p = 0

16 â€“ 8 + 4p = 0

8 + 4p = 0

4p = -8

or p = -2

In the quadratic equation xÂ² + px (1 + 3k) + 7(3 + 2k) = 0

xÂ² â€“ 2x (1 + 3k) + 7(3 + 2k) = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 1, b = -2 (1 + 3k), c = 7 (3 + 2k)

Find Discriminant:

D = bÂ² â€“ 4ac

= (-2(1 + 3k))Â² â€“ 4 x 1 x 7(3 + 2k)

= 4(1 + 9kÂ² + 6k) â€“ 28(3 + 2k)

= 36kÂ² â€“ 32k â€“ 80

Since roots are real and equal, put D = 0

36kÂ² â€“ 32k â€“ 80 = 0

9kÂ² â€“ 8k â€“ 20 = 0

9kÂ² â€“ 18k + 10k â€“ 20 = 0

9k (k â€“ 2) + 10(k â€“ 2) = 0

(k â€“ 2) (9k + 10) = 0

Either, k â€“ 2 = 0 or 9k + 10 = 0

k = 2 or k = -10/9

Question 14:

Solution:

(1 + mÂ²) xÂ² + 2mcx + cÂ² â€“ aÂ² = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (1 + m^2), b = 2mc and c = c^2 â€“ a^2

Since roots are equal, so D = 0

(2mc)2 â€“ 4.(1 + m2)(c2 â€“ a2) = 0

4 m2c2 â€“ 4c2 + 4a2 â€“ 4 m2c2 + 4 m2a2 = 0

a2 + m2a2 = c2

or c2 = a2 (1 + m2)

Hence Proved

Question 15:

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (c^2 â€“ ab) b = â€“ 2(a^2 â€“ bc) c = (b^2 â€“ ac)

Since roots are equal, so D = 0

(â€“ 2(a2 â€“ bc)) 2 â€“ 4(c2 â€“ ab) (b2 â€“ ac) = 0

4(a4 â€“ 2a2bc + b2c2) â€“ 4(b2c2 â€“ ac3 â€“ ab3 + a2bc) = 0

a4 â€“ 3a2bc + ac3 + ab3 = 0

a (a3 â€“ 3abc + c3 + b3) = 0

either a = 0 or (a3 â€“ 3abc + c3 + b3) = 0

a = 0 or a3 + c3 + b3 = 3abc

Hence Proved.

Question 16:

Solution:

2xÂ² + px + 8 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 2, b = p, c = 8

Find D:

D = b2 â€“ 4ac

= pÂ² â€“ 4 x 2 x 8

= pÂ² â€“ 64

Since roots are real, so D â‰¥ 0

pÂ² â€“ 64 â‰¥ 0

pÂ² â‰¥ 64

â‰¥ (Â±8)Â²

Either p â‰¥ 8 or p â‰¤ -8

Question 17:

Solution:

(Î± â€“ 12) xÂ² + 2(Î± â€“ 12) x + 2 = 0

Roots of given equation are equal ( given)

So, D = 0

4(Î± â€“ 12) (Î± â€“ 14) = 0

Î± â€“ 14 = 0 {(Î± â€“ 12) â‰  0}

Î± = 14

Hence the value of Î± is 14

Question 18:

Solution:

9xÂ² + 8kx + 16 = 0

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = 9, b = 8k, c = 16

Find D:

D = bÂ² â€“ 4ac

= (8k)Â² â€“ 4 x 9 x 16

= 64kÂ² â€“ 576

Roots of given equation are equal ( given)

So, D = 0

64kÂ² â€“ 576 = 0

64kÂ² = 576

kÂ² = 9

k = Â±3

Answer: k = 3, k = -3

Question 19:

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

(i) a = k, b = 6, c = 1

For real and distinct roots, then D > 0

6^2 â€“ 4k > 0

36 â€“ 4k > 0

k < 9

(ii)

a = 1, b = â€“ k, c = 9

For real and distinct roots, then D > 0

(-k)^2 â€“ 36 > 0

k > 6 or k < -6

(iii)

a = 9 ,b = 3k ,c = 4

For real and distinct roots, then D > 0

(3k)^2 â€“ 144 > 0

9k^2 > 144

k^2 > 16

k > 4 or k < â€“ 4

(iv)

a = 5, b = â€“ k, c = 1

For real and distinct roots, then D > 0

k^2 â€“ 20 > 0

k2 > 20

k > 2âˆš5 or k < â€“2âˆš5

Question 20:

Solution:

Compare given equation with the general form of quadratic equation, which is ax^2 + bx + c = 0

a = (a â€“ b), b = 5(a + b), c = â€“ 2(a â€“ b)

Find Discriminant:

D = b^2 â€“ 4ac

= (5(a + b))^2 â€“ 4(a â€“ b)(â€“ 2(a â€“ b))

= 25(a + b)^2 + 8(a â€“ b)^2

= 17(a + b)^2 + {8(a + b)^2 + 8(a â€“ b)^2 }

= 17(a + b)^2 + 16(a^2 + b^2)

Which is always greater than zero.

The equation has real and unequal roots.

## R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations

In this chapter students will study important concepts on quadratic equations as listed below:

• Relationship between the roots
• How to find roots of a quadratic equation
• Solving quadratic equations using different methods (Solving quadratic equations by factorisation)
• Solving quadratic equations by completing the square method
• Finding nature of quadratic equations roots using discriminant
• Nature of roots of a quadratic equation

### Key Features of RS Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations

1. R S Aggarwal Solutions consist of a set of solutions to all the important questions.

2. Helps students to know the nature of roots and find roots of any equation at their own pace.

3. Step by step solving approach helps students clear their concepts on quadratic equations.

4. Easy for quick revision.

5. This study material prepared is based on the latest CBSE syllabus by subject experts.