**Solving A Quadratic Equation By Factoring**

Since, 3x^{2} – 5x + 2 is a quadratic polynomial;

3x^{2} – 5x + 2 = 0 is a quadratic equation.

Also,

3x^{2} – 5x + 2 = 3x^{2} – 3x – 2x + 2 [Factorising]

= 3x (x – 1) – 2(x – 1)

= (x – 1) (3x – 2)

In the same way :

3x^{2} – 5x + 2 = 0 ⇒ 3x^{2} – 3x – 2x + 2 = 0 [Factorising L.H.S.]

⇒ (x – 1) (3x – 2) = 0

i.e., x – 1 = 0 or 3x – 2 = 0

⇒ x = 1 or x = 2/3

which is the solution of given quadratic equation.

**In order to solve the given Quadratic Equation:**

1. Clear the fractions and brackets, if given.

2. By transfering each term to the left hand side; express the given equation as

ax^{2 }+ bx + c = 0 or a + bx + cx^{2} = 0

3. Factorise left hand side of the equation obtained (the right hand side being zero).

4. By putting each factor equal to zero; solve it.

**Solving A Quadratic Equation By Factoring With Examples**

**Example 1:** Solve (i) x^{2} + 3x – 18 = 0 (ii) (x – 4) (5x + 2) = 0

(iii) 2x^{2} + ax – a2 = 0; where ‘a’ is a real number.

**Sol. (i)** x^{2} + 3x – 18 = 0

⇒ x^{2} + 6x – 3x – 18 = 0

⇒ x(x + 6) – 3(x + 6) = 0

i.e., (x + 6) (x – 3) = 0

⇒ x + 6 = 0 or x – 3 = 0

⇒ x = – 6 or x = 3

Roots of the given equation are – 6 and 3

**(ii)** (x – 4) (5x + 2) = 0

⇒ x – 4 = 0 or 5x + 2 = 0

x = 4 or x = – 2/5

**(iii)** 2x^{2} + ax – a^{2} = 0

⇒ 2x^{2 }+ 2ax – ax – a^{2 }= 0

⇒ 2x(x + a) – a(x + a) = 0

i.e., (x + a) (2x – a) = 0

⇒ x + a = 0 or 2x – a = 0

⇒ x = – a or x = a/2

**Example 2:** Solve the following quadratic equations

(i) x^{2} + 5x = 0 (ii) x^{2} = 3x (iii) x^{2} = 4

**Sol. (i) **x^{2} + 5x = 0 ⇒ x(x + 5) = 0

⇒ x = 0 or x + 5 = 0

⇒ x = 0 or x = – 5

**(ii) **x^{2} = 3x

⇒ x^{2} – 3x = 0

⇒ x(x – 3) = 0

⇒ x = 0 or x = 3

**(iii)** x^{2} = 4

⇒ x = ± 2

**Example 3:** Solve the following quadratic equations

(i) 7x^{2} = 8 – 10x (ii) 3(x^{2} – 4) = 5x (iii) x(x + 1) + (x + 2) (x + 3) = 42

**Sol. (i) **7x^{2} = 8 – 10x

⇒ 7x^{2} + 10x – 8 = 0

⇒ 7x^{2} + 14x – 4x – 8 = 0

⇒ 7x(x + 2) – 4(x + 2) = 0

⇒ (x + 2) (7x – 4) = 0

⇒ x + 2 = 0 or 7x – 4 = 0

⇒ x = – 2 or x = 4/7

**(ii) **3(x^{2} – 4) = 5x

⇒ 3x^{2} – 5x – 12 = 0

⇒ 3x^{2} – 9x + 4x –¬ 12 = 0

⇒ 3x(x – 3) + 4(x – 3) = 0

⇒ (x – 3) (3x + 4) = 0

⇒ x – 3 = 0 or 3x + 4 = 0

⇒ x = 3 or x = –4/3

**(iii) **x(x + 1) + (x + 2) (x + 3) = 42

⇒ x^{2} + x + x^{2} + 3x + 2x + 6 – 42 = 0

⇒ 2x^{2} + 6x – 36 = 0

⇒ x^{2} + 3x – 18 = 0

⇒ x^{2} + 6x – 3x – 18 = 0

⇒ x(x + 6) – 3(x + 6) = 0

⇒ (x + 6) (x – 3) = 0

⇒ x = – 6 or x = 3

**Example 4:** Solve for x : 12abx^{2} – (9a^{2} – 8b^{2}) x – 6ab = 0

Given equation is 12abx^{2} – (9a^{2} – 8b^{2}) x – 6ab = 0

⇒ 3ax(4bx – 3a) + 2b(4bx – 3a) = 0

⇒ (4bx – 3a) (3ax + 2b) = 0

⇒ 4bx – 3a = 0 or 3ax + 2b = 0

⇒ x =3a/4b or x = – 2b/3a