Chapter 5 - Quadratic Equations Exercise Ex. 5(A)

Question 1

Solution 1

(i) (3x - 1)² = 5(x + 8)

⇒ (9x² - 6x + 1) = 5x + 40

⇒ 9x² - 11x - 39 =0; which is of the form ax² + bx + c = 0.

∴ Given equation is a quadratic equation.

(ii) 5x² - 8x = -3(7 - 2x)

⇒ 5x² - 8x = 6x - 21

⇒ 5x² - 14x + 21 =0; which is of the form ax² + bx + c = 0.

∴ Given equation is a quadratic equation.

(iii) (x - 4)(3x + 1) = (3x - 1)(x +2)

⇒ 3x² + x - 12x - 4 = 3x² + 6x - x - 2

⇒ 16x + 2 =0; which is not of the form ax² + bx + c = 0.

∴ Given equation is not a quadratic equation. 

(iv) X² + 5x – 5 = (x-3)²

x² + 5x – 5 = x² – 6x + 9

11x – 14 = 0; which is not of the form ax² + bx + c = 0

 Given equation is not a quadratic equation

(v) 7x³ - 2x² + 10 = (2x - 5)²

⇒ 7x³ - 2x² + 10 = 4x² - 20x + 25

⇒ 7x³ - 6x² + 20x - 15 = 0; which is not of the form ax² + bx + c = 0.

∴ Given equation is not a quadratic equation.

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

⇒ x² - 2x + 1 + x² + 4x + 4 + 3x + 3 = 0

⇒ 2x² + 5x + 8 = 0; which is of the form ax² + bx + c = 0.

∴ Given equation is a quadratic equation.

 

Question 2

(i) Is x = 5 a solution of the quadratic equation x² - 2x - 15 = 0?

Solution (i)

x² - 2x - 15 = 0

For x = 5 to be solution of the given quadratic equation it should satisfy the equation.

So, substituting x = 5 in the given equation, we get

L.H.S = (5)² - 2(5) - 15

 = 25 - 10 - 15

 = 0

 = R.H.S

Hence, x = 5 is a solution of the quadratic equation x² - 2x - 15 = 0.

(ii) Is x = -3 a solution of the quadratic equation 2x² - 7x + 9 = 0?

Solution (ii)

2x² - 7x + 9 = 0

For x = -3 to be solution of the given quadratic equation it should satisfy the equation

So, substituting x = 5 in the given equation, we get

L.H.S=2(-3)² - 7(-3) + 9

 = 18 + 21 + 9    

 = 48 

 ≠ R.H.S

Hence, x = -3 is not a solution of the quadratic equation 2x² - 7x + 9 = 0. 

Question 3

If  is a solution of equation 3x² + mx + 2 = 0, find the value of m.

Solution 3

For x =  to be solution of the given quadratic equation it should satisfy the equation

So, substituting x =  in the given equation, we get

Question 4

 and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.

Solution 4

For x =  and x = 1 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x =  and x = 1 in the given equation, we get

Solving equations (1) and (2) simultaneously,

Question 5

If 3 and -3 are the solutions of equation ax² + bx - 9 = 0. Find the values of a and b.

Solution 5

For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = 3 and x = -3 in the given equation, we get

Solving equations (1) and (2) simultaneously,