# Quadratic Equations – Exercise 9.2 – Class X

## Previous Exercise – Quadratic Equations – Exercise 9.1 – Class X

### Quadratic equations – Exercise 9.2

(i) x2 = 100

(ii) x2 + 6 = 100

(iii) p(p – 3) = 1

(iv) x3 + 3 = 2x

(v) (x+ 9)(x – 9) = 0

(vi) 2x2 = 72

(vii) x2 – x = 0

(viii)7x = 35/x

(ix) x + 1/x = 5

(x) 4x = 81/x

(xi) (2x – 5)2 = 81

(xii) (x – 4)^2/18 = 2/9

(i)x2 – 196 = 0

(ii) 5x2= 625

(iii) x2 + 1 = 101

(iv) 7x = 64/7x

(v) (x + 8)2 – 5=31

(vi) x^2/23/4 = 71/4

(vii) -4x2 + 324 = 0

(viii) -37.5 x2 = -37.5

3.In each of the following determine whether the given values of variables is a solution of the quadratic equation or not.

(i)x2 + 14x + 13 = 0; x = -1 ; x =-13

(ii) 7x2 – 12x = 0 ; x = 1/3

(iii) 2m2 – 6m + 3 = 0; m = 1/2

(iv) y2 + √2y – 4 = 0 ; y = 2√2

(v) x/x+2 = 1/2 ; x = 2 and x = 1

(vi) 6x2 – x – 2 = 0 ; x = –1/2 and x = 2/3

(i)If A= πr2, solve for r and find the value of r if A = 77 and π = 22/7

(ii) If r2 = l2 + d2 solve for d and find the value of d if r = 5 and l = 4

(iii) If A = √3a^2/4 solve for v and find the value of a, if A = 16√3

(iv) if k = 1/2 mv2 solve for v and find the value of v, if k = 100 and m = 2

(vi) If  v2 = u2 + 2as solve for v and find the value of v, if u = 0, a = 2, s = 100

## Quadratic equations – Exercise 9.2

(i) x2 = 100

Solution:

The given equation, x2 = 100

x2 – 100 = 0

The given equation is having only second degree variable.

Therefore, the given equation x2 = 100 is pure quadratic equation.

(ii) x2 + 6 = 100

Solution:

The given equation, x2 + 6 = 100

x2 + 6 – 100 = 0

x2 – 94 = 0

The given equation is having only second degree variable.

Therefore, the given equation x2 + 6 = 100 is pure quadratic equation.

(iii) p(p – 3) = 1

Solution:

The given equation, p(p – 3) = 1

p2 – 3p – 1 = 0

The given equation is having both second degree and first degree variable.

Therefore, the given equation p(p – 3) = 0 is adfected quadratic equation.

(iv) x3 + 3 = 2x

Solution:

The given equation, x3 + 3= 2x

x3 – 2x + 3 = 0

The given equation is having third degree variable.

Therefore, the given equation x3 + 3 = 2x is not a quadratic equation.

(v) (x+ 9)(x – 9) = 0

Solution:

The given equation, (x + 9)(x – 9) = 0

x2 – 9x + 9x – 81 = 0

x2 – 81 = 0

The given equation is having only second degree variable.

Therefore, the given equation (x + 9)(x – 9)= 0 is pure quadratic equation.

(vi) 2x2 = 72

Solution:

The given equation, 2x2 = 72

2x2 – 72 = 0

The given equation is having only second degree variable.

Therefore, the given equation 2x2 = 72 is pure quadratic equation.

(vii) x2 – x = 0

Solution:

The given equation, x2 – x = 0

The given equation is having both second degree and first degree variable.

Therefore, the given equation x2 – x = 0 is adfected quadratic equation.

(viii)7x = 35/x

Solution:

The given equation, 7x = 35/x

7x2 = 35

7x2 – 35 = 0

The given equation is having only second degree variable.

Therefore, the given equation 7x = 35/x is pure quadratic equation.

(ix) x + 1/x = 5

Solution:

The given equation, x + 1/x = 5

x2 + 1 = 5x

x2 – 5x + 1 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation x2 – 5x + 1 = 0 is adfected quadratic equation.

(x) 4x = 81/x

Solution:

The given equation, 4x = 81/x

4x2 = 81

4x2 – 81 = 0

The given equation is having only second degree variable.

Therefore, the given equation 4x = 81/x is pure quadratic equation.

(xi) (2x – 5)2 = 81

Solution:

The given equation, (2x – 5)2 = 81

(2x)2 – 2(2x)(5) + 52 = 81

4x2 – 10x + 25 – 81 = 0

4x2 – 10x – 56 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation (2x – 5)2 = 81 is adfected quadratic equation.

(xii) (x – 4)^2/18 = 2/9

Solution:

The given equation, (x – 4)^2/18 = 2/9

9(x – 4)2 = 2 x 18

9(x2 – 8x + 16) = 36

9x2 – 72x + 144 – 36 = 0

9x2 – 72x + 108 = 0

The given equation is having both first degree and second degree variable.

Therefore, the given equation (x – 4)^2/18 = 2/9 is adfected quadratic equation.

(i)x2 – 196 = 0

Solution:

x2 – 196 = 0

x2 = 196

x = √196

x = ±14

(ii) 5x2= 625

Solution:

5x2 = 625

x2 = 625/5 = 125

x = √125 = ±5√5

(iii) x2 + 1 = 101

Solution:

x2 = 101 – 1

x2 = 100

x = √100 = ±10

(iv) 7x = 64/7x

Solution:

7x = 64/7x

49x2 = 64

x2 = 64/49

x = √(64/49)

x = ±8/7

(v) (x + 8)2 – 5=31

Solution:

(x + 8)2 – 5 = 31

(x + 8)2 = 31 + 5

(x + 8)2 = 36

(x + 8) = √36

x + 8 = ±6

x + 8 = 6 or x + 8 = -6

x = 6 – 8 or x = -6 – 8

x = -2 or x = -14

(vi) x^2/23/4 = 71/4

Solution:

x^2/23/4 = 71/4

x^2/23/4 = 29/4

2x2 – 3 = 29

2x2 = 29 + 3

2x2 = 32

x2 = 32/2 = 16

x = √16

x = ±4

(vii) -4x2 + 324 = 0

Solution:

-4x2 + 324 = 0

-4x2 = -324

4x2 = 324

x2 = 324/4 = 81

x = √81 = ±9

(viii) -37.5 x2 = -37.5

Solution:

-37.5 x2 = -37.5

x2 = -37.5/-37.5 = 1

x = √1 = ±1

3.In each of the following determine whether the given values of variables is a solution of the quadratic equation or not.

(i) x2 + 14x + 13 = 0; x = -1 ; x =-13

Solution:

If x = -1 then x2 + 14x + 13 = 0

x2 + 14x + 13 = (-1)2 + 14(-1) + 13

x2 + 14x + 13 = 1 – 14 + 13

x2 + 14x + 13 = 14 – 14

x2 + 14x + 13 = 0

Therefore x = -1 is the solution of the equation x2 + 14x + 13 = 0

Given, if x = -13 then x2 + 14x + 13

x2 + 14x + 13 = (-13)2 + 14(-13) + 13

x2 + 14x + 13 = 196  – 182 + 13

x2 + 14x + 13 = 27

if x = -13 then x2 + 14x + 13 ≠ 0

Therefore x = -13 is not the solution of the equation x2 + 14x + 13 = 0

(ii) 7x2 – 12x = 0 ; x = 1/3

Solution:

Given, if x = 1/3 then 7x2 – 12x = 0

7x2 – 12x = 7(1/3)2 – 12(1/3)

7x2 – 12x = 7/912/3

7x2 – 12x = 7/9 – 4 = –29/9

If x = 1/3 then 7x2 – 12x ≠ 0

Therefore x = 1/3 is not the solution of the equation 7x2 – 12x = 0

(iii) 2m2 – 6m + 3 = 0; m = 1/2

Solution:

Given, if m = 1/2 then 2m2 – 6m + 3 = 0

2m2 – 6m + 3 = 2(1/2)2 – 6(1/2) + 3

2m2 – 6m + 3 = 21/4 – 2 + 3

2m2 – 6m + 3 = 1/2

If x = 1/2 then 2m2 – 6m + 3 ≠ 0

Therefore x = 1/2 is not the solution of the equation 2m2 – 6m + 3 = 0

(iv) y2 + 2y – 4 = 0 ; y = 2√2

Solution:

Given, if y = (2√2) then y2 + √2y – 4 = 0

y2 + √2y – 4 = (2√2)2 + √2(2√2)y – 4

y2 + √2y – 4 = 4×2 + 2×2 – 4

y2 + √2y – 4 = 8 + 4 – 4

y2 + √2y – 4 = 8

If y = 2√2 then y2 + √2y – 4 ≠ 0

Therefore y = (2√2) is not the solution of the equation y2 + √2y – 4 = 0

(v) x/x+2 = 1/2 ; x = 2 and x = 1

Solution:

Given, if x = 2 then x/x+21/2 = 0

x/x+21/2 = 2/2 + 2  – 1/2 = 2/41/2 = 1/21/2 = 0

Therefore x = 2 is the solution of the equation x/x+21/2 = 0

Given, if x = 1 then x/x+21/2 = 0

x/x+21/2 = 1/1 + 2  – 1/2 = 1/31/2 = 2 – 3 /6  = -1/6

Therefore x = 1 is not the solution of the equation x/x+21/2 = 0

(vi) 6x2 – x – 2 = 0 ; x = –1/2 and x = 2/3

Given, if x = (-1/2)  then  6x2 – x – 2 = 0

6x2 – x – 2 = 6(-1/2)2 – (-1/2) – 2

6x2 – x – 2 = 6(1/4) + 1/2 – 2

6x2 – x – 2 = 3/2 + 1/2 – 2

6x2 – x – 2 = 2 – 2

6x2 – x – 2 = 0

Therefore x = 2 is the solution of the equation  6x2 – x – 2 = 0

(i)If A= πr2, solve for r and find the value of r if A = 77 and π = 22/7

Solution:

A= πr2

77 = 22/7 x r2

77 x 7 = 22 r2

539 = 22r2

539/22 = r2

49/2 = r2

r = ±7/√2

(ii) If r2 = l2 + d2 solve for d and find the value of d if r = 5 and l = 4

Solution:

r2 = l2 + d2

52 = 42 + d2

d2 = 52 – 42

d2 = 25 – 16 = 9

d = ±3

(iii) If A = √3a^2/4 solve for v and find the value of a, if A = 16√3

Solution:

16√3 = √3a^2/4

16√3 x 4 = √3a2

a2 = 16√3 x 4/√3

a2 = 16 x 4 = 64

a = ±8

(iv) if k = 1/2 mv2 solve for v and find the value of v, if k = 100 and m = 2

Solution:

k = 1/2 mv2

100 = 1/2 x 2 x v2

100 x 2 = 2v2

200/2 = v2

100 = v2

v = ±10

(vi) If  v2 = u2 + 2as solve for v and find the value of v, if u = 0, a = 2, s = 100

Solution:

v2 = u2 + 2as

v2 = 0 + 2(2)(100)

v2 = 400

v2 = √400

v = ±20

Next  Exercise – Quadratic Equations – Exercise 9.3 – Class X