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

### Quadratic equations – Exercise 9.2

1.Classify the following equations into pure and adfected quadratic equations:

(i) x^{2} = 100

(ii) x^{2} + 6 = 100

(iii) p(p – 3) = 1

(iv) x^{3} + 3 = 2x

(v) (x+ 9)(x – 9) = 0

(vi) 2x^{2} = 72

(vii) x^{2} – 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}

2.Solve the quadratic equations:

(i)x^{2} – 196 = 0

(ii) 5x^{2}= 625

(iii) x^{2} + 1 = 101

(iv) 7x = ^{64}/_{7x}

(v) (x + 8)^{2} – 5=31

(vi) ^{x^2}/_{2} – ^{3}/_{4} = 7^{1}/_{4}

(vii) -4x^{2} + 324 = 0

(viii) -37.5 x^{2} = -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)x^{2} + 14x + 13 = 0; x = -1 ; x =-13

(ii) 7x^{2 }– 12x = 0 ; x = ^{1}/_{3}

(iii) 2m^{2} – 6m + 3 = 0; m = ^{1}/_{2}

(iv) y^{2} + √2y – 4 = 0 ; y = 2√2

(v) ^{x}/_{x+2} = ^{1}/_{2} ; x = 2 and x = 1

(vi) 6x^{2} – x – 2 = 0 ; x = –^{1}/_{2} and x = ^{2}/_{3}

4. Solve the quadratic equations:

(i)If A= πr^{2}, solve for r and find the value of r if A = 77 and π = ^{22}/_{7}

(ii) If r^{2} = l^{2} + d^{2 }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} mv^{2} solve for v and find the value of v, if k = 100 and m = 2

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

## Quadratic equations – Exercise 9.2

**1.Classify the following equations into pure and adfected quadratic equations:**

**(i) x ^{2} = 100**

Solution:

The given equation, x^{2} = 100

x^{2} – 100 = 0

The given equation is having only second degree variable.

Therefore, the given equation x^{2} = 100 is pure quadratic equation.

** (ii) x ^{2} + 6 = 100**

Solution:

The given equation, x^{2} + 6 = 100

x^{2} + 6 – 100 = 0

x^{2} – 94 = 0

The given equation is having only second degree variable.

Therefore, the given equation x^{2} + 6 = 100 is pure quadratic equation.

** (iii) p(p – 3) = 1**

Solution:

The given equation, p(p – 3) = 1

p^{2} – 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) x ^{3} + 3 = 2x**

Solution:

The given equation, x^{3} + 3= 2x

x^{3} – 2x + 3 = 0

The given equation is having third degree variable.

Therefore, the given equation x^{3} + 3 = 2x is not a quadratic equation.

** (v) (x+ 9)(x – 9) = 0**

Solution:

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

x^{2} – 9x + 9x – 81 = 0

x^{2} – 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) 2x ^{2} = 72**

Solution:

The given equation, 2x^{2} = 72

2x^{2} – 72 = 0

The given equation is having only second degree variable.

Therefore, the given equation 2x^{2} = 72 is pure quadratic equation.

** (vii) x ^{2} – x = 0**

Solution:

The given equation, x^{2} – x = 0

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

Therefore, the given equation x^{2} – x = 0 is adfected quadratic equation.

** (viii)7x = ^{35}/_{x}**

Solution:

The given equation, 7x = ^{35}/_{x}

7x^{2} = 35

7x^{2} – 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

x^{2} + 1 = 5x

x^{2} – 5x + 1 = 0

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

Therefore, the given equation x^{2} – 5x + 1 = 0 is adfected quadratic equation.

** (x) 4x = ^{81}/_{x}**

Solution:

The given equation, 4x = ^{81}/_{x}

4x^{2} = 81

4x^{2} – 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) + 5^{2} = 81

4x^{2} – 10x + 25 – 81 = 0

4x^{2} – 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(x^{2} – 8x + 16) = 36

9x^{2} – 72x + 144 – 36 = 0

9x^{2} – 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.

**2.Solve the quadratic equations:**

**(i)x ^{2} – 196 = 0**

Solution:

x^{2} – 196 = 0

x^{2} = 196

x = √196

x = ±14

** (ii) 5x ^{2}= 625**

Solution:

5x^{2} = 625

x^{2} = ^{625}/_{5} = 125

x = √125 = ±5√5

** (iii) x ^{2} + 1 = 101**

Solution:

x^{2} = 101 – 1

x^{2} = 100

x = √100 = ±10

** (iv) 7x = ^{64}/_{7x}**

Solution:

7x = ^{64}/_{7x}

49x^{2} = 64

x^{2} = ^{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}/_{2} – ^{3}/_{4} = 7^{1}/_{4}**

Solution:

^{x^2}/_{2} – ^{3}/_{4} = 7^{1}/_{4}

^{x^2}/_{2} – ^{3}/_{4} = ^{29}/_{4}

2x^{2} – 3 = 29

2x^{2} = 29 + 3

2x^{2} = 32

x^{2} = ^{32}/_{2} = 16

x = √16

x = ±4

** (vii) -4x ^{2} + 324 = 0**

Solution:

-4x^{2} + 324 = 0

-4x^{2} = -324

4x^{2} = 324

x^{2} = ^{324}/_{4} = 81

x = √81 = ±9

** (viii) -37.5 x ^{2} = -37.5**

Solution:

-37.5 x^{2} = -37.5

x^{2} = ^{-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) x ^{2} + 14x + 13 = 0; x = -1 ; x =-13**

Solution:

If x = -1 then x^{2} + 14x + 13 = 0

x^{2} + 14x + 13 = (-1)^{2} + 14(-1) + 13

x^{2} + 14x + 13 = 1 – 14 + 13

x^{2} + 14x + 13 = 14 – 14

x^{2} + 14x + 13 = 0

Therefore x = -1 is the solution of the equation x^{2} + 14x + 13 = 0

Given, if x = -13 then x^{2} + 14x + 13

x^{2} + 14x + 13 = (-13)^{2} + 14(-13) + 13

x^{2} + 14x + 13 = 196 – 182 + 13

x^{2} + 14x + 13 = 27

if x = -13 then x^{2} + 14x + 13 ≠ 0

Therefore x = -13 is not the solution of the equation x^{2} + 14x + 13 = 0

** (ii) 7x ^{2 }– 12x = 0 ; x = ^{1}/_{3}**

Solution:

Given, if x = ^{1}/_{3} then 7x^{2} – 12x = 0

7x^{2} – 12x = 7(^{1}/_{3})^{2} – 12(^{1}/_{3})

7x^{2} – 12x = ^{7}/_{9} – ^{12}/_{3}

7x^{2} – 12x = ^{7}/_{9} – 4 = –^{29}/_{9}

If x = ^{1}/_{3} then 7x^{2} – 12x ≠ 0

Therefore x = ^{1}/_{3} is not the solution of the equation 7x^{2} – 12x = 0

** (iii) 2m ^{2} – 6m + 3 = 0; m = ^{1}/_{2}**

Solution:

Given, if m = ^{1}/_{2} then 2m^{2} – 6m + 3 = 0

2m^{2} – 6m + 3 = 2(^{1}/_{2})^{2} – 6(^{1}/_{2}) + 3

2m^{2} – 6m + 3 = 2^{1}/_{4} – 2 + 3

2m^{2} – 6m + 3 = ^{1}/_{2}

If x = ^{1}/_{2} then 2m^{2} – 6m + 3 ≠ 0

Therefore x = ^{1}/_{2} is not the solution of the equation 2m^{2} – 6m + 3 = 0

** (iv) y ^{2} + **

**√**

**2y – 4 = 0 ; y = 2**

**√2**

Solution:

Given, if y = (2√2) then y^{2} + √2y – 4 = 0

y^{2} + √2y – 4 = (2√2)^{2} + √2(2√2)y – 4

y^{2} + √2y – 4 = 4×2 + 2×2 – 4

y^{2} + √2y – 4 = 8 + 4 – 4

y^{2} + √2y – 4 = 8

If y = 2√2 then y^{2} + √2y – 4 ≠ 0

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

** (v) ^{x}/_{x+2} = ^{1}/_{2} ; x = 2 and x = 1**

Solution:

Given, if x = 2 then ^{x}/_{x+2} – ^{1}/_{2} = 0

^{x}/_{x+2} – ^{1}/_{2} = ^{2}/_{2 + 2 } – ^{1}/_{2} = ^{2}/_{4} – ^{1}/_{2} = ^{1}/_{2} – ^{1}/_{2} = 0

Therefore x = 2 is the solution of the equation ^{x}/_{x+2} – ^{1}/_{2} = 0

Given, if x = 1 then ^{x}/_{x+2} – ^{1}/_{2} = 0

^{x}/_{x+2} – ^{1}/_{2} = ^{1}/_{1 + 2 } – ^{1}/_{2} = ^{1}/_{3} – ^{1}/_{2} = ^{2 – 3 }/_{6} = ^{-1}/_{6}

Therefore x = 1 is not the solution of the equation ^{x}/_{x+2} – ^{1}/_{2} = 0

** (vi) 6x ^{2} – x – 2 = 0 ; x = –^{1}/_{2} and x = ^{2}/_{3}**

Given, if x = (-^{1}/_{2}) then 6x^{2} – x – 2 = 0

6x^{2} – x – 2 = 6(-^{1}/_{2})^{2} – (-^{1}/_{2}) – 2

6x^{2} – x – 2 = 6(^{1}/_{4}) + ^{1}/_{2} – 2

6x^{2} – x – 2 = ^{3}/_{2} + ^{1}/_{2} – 2

6x^{2} – x – 2 = 2 – 2

6x^{2} – x – 2 = 0

Therefore x = 2 is the solution of the equation 6x^{2} – x – 2 = 0

**4.Solve the quadratic equations:**

**(i)If A= πr ^{2}, solve for r and find the value of r if A = 77 and π = ^{22}/_{7}**

Solution:

A= πr^{2}

77 = ^{22}/_{7} x r^{2}

77 x 7 = 22 r^{2}

539 = 22r^{2}

^{539}/_{22} = r^{2}

^{49}/_{2} = r^{2}

r = ±^{7}/_{√2}

** (ii) If r ^{2} = l^{2} + d^{2 }solve for d and find the value of d if r = 5 and l = 4**

Solution:

r^{2} = l^{2} + d^{2}

5^{2} = 4^{2} + d^{2}

d^{2} = 5^{2} – 4^{2}

d^{2} = 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 = √3a^{2}

a^{2} = ^{16√3 x 4}/_{√3}

a^{2} = 16 x 4 = 64

a = ±8

** (iv) if k = ^{1}/_{2} mv^{2} solve for v and find the value of v, if k = 100 and m = 2**

Solution:

k = ^{1}/_{2} mv^{2}

100 = ^{1}/_{2} x 2 x v^{2}

100 x 2 = 2v^{2}

^{200}/_{2} = v^{2}

100 = v^{2}

v = ±10

** (vi) If v ^{2} = u^{2} + 2as solve for v and find the value of v, if u = 0, a = 2, s = 100**

Solution:

v^{2} = u^{2} + 2as

v^{2} = 0 + 2(2)(100)

v^{2} = 400

v^{2} = √400

v = ±20

**Next Exercise – Quadratic Equations – Exercise 9.3 – Class X**