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
