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Class 10 Quadratic Equations Additional Questions

Topic– 3: Solution of Quadratic Equations by Completing the Square

 Short Answer Type Questions: [2 Marks]

1. Find the roots of the following equation 4x2 + 4bx – (a2 – b2) = 0 by the method of completing the square.

Sol. We have, 4x2 + 4bx – (a2 – b2) = 0

2. Solve 2x2 + 14x + 9 = 0, by completing the squares, when (i) x is a rational number, (ii) x is a real number.

Sol. Given equation can be rewritten as

⇒ 2x2 + 14x = –9

⇒ x2 + 7x = − 9/2

which are both irrational.

Thus, (i) when x is a rational number, the equation has no solution.

(ii) when x is a real number, the roots are

Short Answer Type Questions: [3 Marks]

1. Using the quadratic formula, solve the equation :

2. Solve each of the following equations by using quadratic formula :

(i) a(x2 + 1) = x(a2 + 1)

(ii) x2 + 2ab = (2a + b)x

(iii) 3y2 + (6 + 4a)y + 8a = 0

Sol. (i) We have, a(x2 + 1) = x(a2 + 1)

⇒ ax2 – x(a2 + 1) + a = 0, which is quadratic equation in x.

Here, A = a, B = –(a2 + 1) and C = a

Using quadratic formula, we get

ii) We have, x2 + 2ab = (2a + b)x

⇒ x2 – (2a + b)x + 2ab = 0

Here, A = 1, B = – (2a + b), C = 2ab

B2 – 4AC = [–(2a + b)]2 – 4 × 1 × 2ab

= (2a + b)2 – 8ab

= 4a2 + b2 + 4ab – 8ab

= 4a2 + b2 – 4ab = (2a – b)2

Using quadratic formula, we get

Hence, x = 2a, b.

(iii) We have, 3y2 + (6 + 4a)y + 8a = 0

Here, A = 3, B = 6 + 4a, C = 8a

B2 – 4AC = (6 + 4a)2 – 4 × 3 × 8a

= 36 + 16a2 + 48a – 96a

= 36 + 16a2 – 48a = (6 – 4a)2

Using quadratic formula, we get

3. Using quadratic formula, solve the following quadratic equation for x :

Long Answer Type questions: [4 Marks]

Using quadratic formula, solve for

2. Using quadratic formula, solve the following for x : abx2+ (b2−ac)x −bc =0

Sol. We have, abx2+ (b2−ac)x −bc =0

On comparing it with Ax2 + Bx + C + =0 , we get :

A = ab, B = b2 – ac, C = –bc

⇒   3 x + 6 = x2 – x – 2

⇒ x2 – 4x – 8 = 0

Here, a = 1, b = – 4 and c = – 8

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