**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 4x**^{2}** + 4bx – (a**^{2}** – b**^{2}**) = 0 by the method of completing the square. **

**Sol. **We have, 4x^{2} + 4bx – (a^{2} – b^{2}) = 0

**2. **Solve 2x^{2 }+ 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

⇒ 2x^{2} +
14x = –9

⇒ x^{2} + 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(x^{2} + 1) = x(a^{2} + 1)

(ii) x^{2} + 2ab = (2a + b)x

(iii) 3y^{2} + (6 + 4a)y + 8a = 0

**Sol.
**(i) We have, a(x^{2} + 1) = x(a^{2} + 1)

⇒ ax^{2} – x(a^{2} + 1) + a =
0, which is quadratic equation in x.

Here, A = a, B = –(a^{2} + 1) and C = a

Using quadratic formula, we get

ii) We have, x^{2}
+ 2ab = (2a + b)x

⇒ x^{2} –
(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

= 4a^{2} + b^{2} + 4ab – 8ab

= 4a^{2}
+ b^{2} – 4ab = (2a – b)^{2}

Using quadratic formula, we get

Hence, x = 2a, b.

(iii) We have, 3y^{2} + (6 + 4a)y + 8a = 0

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

B^{2} – 4AC
= (6 + 4a)^{2} – 4 × 3
× 8a

= 36 + 16a^{2} + 48a – 96a

= 36 + 16a^{2} – 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
: abx^{2}+ (b^{2}−ac)x −bc =0

**Sol.
**We have, abx^{2}+
(b^{2}−ac)x −bc =0

On comparing it with
Ax^{2} + Bx + C + =0 , we get :

A = ab, B = b^{2} – ac, C = –bc

⇒ 3 x +
6 = x^{2} – x – 2

⇒ x^{2}
– 4x – 8 = 0

Here, a = 1, b = – 4 and c = – 8