#### Solving Quadratic Equations by Factorization

ax2 + bx + c, a ≠ 0, a, b, c ε R is called a quadratic polynomial or quadratic expression in x.

ax2 + b x + c = 0 (a ≠ 0 ) is called a quadratic equation.

We will now try to find the values of x that satisfy ax2 + b x + c = 0, i.e., we will find the solutions of
ax2 + b x + c = 0.

Recall that if a ≠ 0 for two real numbers a, b then either a = 0 or b = 0 or both equal zero.

If a ≠ 0, then if we multiply either side of a b = 0 by 1/a, we get b = 0.

Therefore, a b = 0 implies that either a = 0 or b = 0 or both.

#### Solving Quadratic Equations with the Factorization Method

To apply this method to solve ax2 + b x + c = 0, ax2 + b x + c must be resolvable into factors.

1. Write ax2 + b x + c as a product of its factors.

2. For this product to be zero, one of the factors should be zero. Recall that the value of x for which the value of
a polynomial becomes zero is called a zero of the polynomial. It is also called a solution to the equation.
Therefore, find the zeros of the factors that give solutions to the equation ax2 + b x + c= 0. The solutions
of ax2 + b x + c = 0 are called the roots of the equation.

The following example illustrates the method.

#### Example 1

Solve 2x2 - x - 6 = 0

Solution:

We resolve 2x2 - x - 6 into factors.

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

= 2x ( x - 2 ) + 3 ( x - 2 )

= ( x - 2 ) ( 2x + 3 )

Therefore, the given equation 2x2 - x - 6 = 0 reduces to ( x - 2 ) ( 2x + 3 ) = 0, which implies either x - 2 = 0 or 2x + 3 = 0, i.e., x = 2 or x = - 3/2.

x = 2, - 3/2 are the roots of the equation.

Solution set of the equation = { 2, - 3/2 }

#### Example 2

Solve 1/x + a - 1/ x + b = 1/a - 1/b

Solution:

Determine whether this equation reduces to the quadratic form.

The given equation is

( x + b ) - ( x + a ) / ( x + a ) ( x + b ) = ( b - a )/ab

i.e., ( b - a ) / ( x + a ) ( x + b ) = ( b - a ) / ab

( x + a ) ( x + b ) = ab/p>

i.e., x2 + ( a + b ) x + ab = ab

Therefore, x2 + ( a + b ) x = 0

i.e., x ( x + ( a + b ) ) = 0

i.e., x = 0 or x + ( a + b ) = 0

i.e., x = 0 or x = - ( a + b )

Solution set = { 0, - ( a + b ) }

#### Try these questions

1. 6x2 - 13x + 6 = 0

2. x2 - 2ax + ( a2 - b2 ) = 0

3. x2 + x ( c - b ) + ( c - a ) ( a - b ) = 0

4. ( 2x + 1 ) / ( a + 1 ) + 2a /x = 5

5. ( a - x )2 + ( b - x )2 = ( a - b )2

6. 1/x + 1/x +a = 1/b + 1/ b+ a

7. x /( x - b ) + x = a / ( a - b ) + a

8. a / ( x - b ) + b/( x - a ) = 2

9. ( x2 + ax + a2 ) / ( x2 - ax + a2 ) = 13/7

10. If 3 f(x) = 3 [ 3x2 - 2x + 1] and 4g(x) = 4(2x - 3/2) find x for which 3f(x) = 4g(x)

11. f(x) = 2x2 - 3x - 5 find the value of x if f(3x) = f(2x) + 81

12. f (x) = ( 5x - 4 )2 - 3 ( 2x - 5 )2

13. g(x) = (4x - 3)2 - 2(x - 3)2 + 10 for what value of x is f(x) = g(x) ?

1. 6x2 – 13x + 6 = 0

6x2 – 9x – 4x + 36 = 0

3x ( 2x – 3 ) – 2( 2x – 3 ) = 0

( 2x – 3 ) ( 3x – 2 ) = 0

2x – 3 = 0 or ( 3x – 2 ) = 0

2x = 3 or 3x = 2

x = 3/2 or x = 2/3 ; Therefore x = 3/2 or 2/3.

2. x2 – 2ax + ( a2 – b2 ) = 0

[ a2 – b2 = ( a + b ) ( a – b ) and – ( a + b ) – ( a – b ) = – a –b – a + b = – 2a ]

x2 – ( a + b )x – ( a – b )x + ( a2 – b2 ) = 0

= x [ x – ( a + b ) ] – ( a – b ) [ x – ( a + b ) ] = 0

[ x – ( a + b ) ] [ x – ( a – b ) ] = 0

x – ( a + b ) = 0 or x - ( a – b ) = 0

x = a + b or x = ( a – b )

Therefore x = a + b, a - b

3. x2 + x ( c – b ) + ( c – a ) ( a – b ) = 0

[ c – a + a – b = c – b ]

Therefore x2 + x( c – a ) + x ( a – b ) + ( c – a ) ( a – b ) = 0

Therefore x + x ( c – a ) + x( a – b ) + ( c – a ) ( a – b ) =0

x [ x + ( c – a ) ] + ( a – b ) [ x + ( c – a ) ] = 0

[ x + ( c – a ) ] [ x + ( a – b ) ] = 0

( x + c – a ) ( x + a – b ) = 0

x + c – a = 0 or x + a – b = 0

x = a – c or x = b – a

Therefore x = ( a – c ) or ( b – a ).

4. ( 2x + 1 ) /( a + 1 ) + 2a /x = 5

{x ( 2x + 1 ) + 2a ( a + 1 )} / x ( a + 1 ) = 5

2x2 + x + 2a2 + 2a = 5x ( a + 1 )

2x2 + x + 2a2 + 2a = 5ax + 5x

2x2 + x + 2a2 + 2a – 5ax – 5x = 0

2x2 – x ( 5a + 4 ) + 2a ( a + 1 ) = 0

2x [ x – 2 ( a + 1 ) ] – a [ x – 2 ( a + 1 )] = 0

[ x – 2 ( a + 1 ) ] [ 2x – a ] = 0

x – 2 ( a + 1 ) = 0 or 2x – a = 0

That is x = 2 ( a + 1 ) or 2x = a

x = 2( a + 1 ) or a /2.

5. ( a – x )2 + ( b – x )2 = ( a – b )2

a2 – 2ax + x2 + b2 – 2bx + x2 = a2 – 2ab + b2

2x2 – 2ax – 2bx + 2ab = 0

2x2 – 2x ( a + b ) + 2ab = 0

[ 2 *2ab = 4ab; 4ab = 2a * 2b and (– 2a) + (– 2b) = – ( 2a + 2b )]

2x2 – 2ax – 2bx + 2ab = 0

= 2x ( x – a ) – 2b ( x – a ) = 0

( x – a ) ( 2x – 2b ) =0

x – a = 0 or 2x – 2b = 0

x = a or b.

6. 1/x + 1/( x + a ) = 1/b + 1/( b + a )

1/x – 1/b = 1/( b + a ) – 1/ ( x + a )

( b – x ) / bx = ( x + a – b – a ) / (b +a ) (x + a)

(b – x ) / b x = – (b – x) / (b + a) (x + a)

if b – x = 0, then x = b

if b – x ≠0, canceling ( b – x ) in the numerator on both sides of

( b – x ) / bx = – ( b – x ) / ( b + a ) ( x + a )

we have 1 / b x = – 1/{ ( b + a ) ( x + a ) }

That is, ( b + a ) ( x + a ) = – bx

bx + ab + ax + a2 = – bx

bx + ax + bx = – ab – a2

2bx + ax = – ab – a2

( 2b + a ) x = – a ( b + a )

Therefore x = – a ( b + a ) / 2b + a = – a ( a + b ) / ( a + 2b )

Therefore x = b or – a ( a + b ) / ( a + 2b )

7. x /( x – b ) + x = a / ( a – b ) + a

x / ( x – b ) – a / ( a – b ) = a – x

( ax – b x – ax + ab ) / ( x – b ) ( a – b ) = a – x

b ( a – x ) / { ( x – b ) ( a – b ) } = a – x.

If a – x = 0, then x = a

If a – x ≠0, cancel ( a – x ) from the numerator on both sides,

We get b / ( x – b ) ( a – b ) = 1

( x – b ) ( a – b ) = b

x ( a – b ) – b ( a – b ) = b

x ( a – b ) = b + b ( a – b ) = b + ab – b2

Therefore x = b + ab – b2 / a – b = b ( a – b + 1 )/ ( a – b )

Therefore x = a or b (a – b + 1) / ( a – b ).

8. a /( x – b ) + b/( x – a ) = 2

( ax – a2 + bx – b2 ) / { ( x – b ) ( x – a ) } = 2

ax – a2 + bx – b2 = 2( x –b ) ( x – a )

ax – a2 + bx – b2 = 2( x2 – ax – bx + ab )

ax – a2 + bx – b2 = 2x2 – 2ax – 2bx + 2ab

2x2 – 2ax – 2bx + 2ab – ax – b x + a2 + b2 = 0

2x2 – 3x ( a + b ) + ( a2 + 2ab + b2 ) = 0

2x2 – 3x ( a + b ) + ( a + b )2 = 0

2x2 – 2x( a + b ) – x( a + b ) + ( a + b )2 = 0

2x [ x – ( a + b ) ] – ( a + b ) [ x – ( a + b ) ] = 0

[ x – ( a + b ) ] [ 2x – ( a + b ) ] = 0

x = ( a + b ) or 2x – ( a + b ) = 0

x = ( a + b ) or 2x = a + b

Therefore x = a + b or ( a + b ) /2.

9. ( x2 + ax + a2 ) / ( x2 – ax + a2 ) = 13 / 7

7x2 + 7ax + 7a2 = 13x2 – 13ax + 13a2

13x2 – 13ax + 13a2 – 7x2 – 7ax – 7a2 =0

6x2 – 20ax + 6a2 = 0

3x2 – 10ax + 3a2 = 0

[ 3 * 3a2 = 9a2 ; 9a * a and ( – 9a ) + ( – a ) = – 10a ]

3x2 – 10ax + 3a2 = 0 = 3x2 – 9ax – ax + 3a2

3x ( x – 3a ) – a ( x – 3a ) = 0

( x – 3a ) ( 3x – a ) = 0

x – 3a = 0 or 3x – a = 0

x = 3a or 3x = a

Therefore x = 3a or a/3.

10. 3 f(x) = 3 [ 3x2 – 2x + 1]

= 9x2 – 6x + 3

4g (x) = 4 [ 2x – 3/2] = 8x2 – 6

3 f(x) = 4g (x)

9x2 – 6x + 3 = 8x2 – 6

9x2 – 6x + 3 – 8x2 + 6 = 0

x2 – 6x + 9 = 0

( x – 3 )2 = 0 or ( x – 3 ) =0

x = 3

if x = 3, then 3 f(x) = 4g (x)

11. f(x) = 2x2 – 3x – 5

f(2x) = 2 ( 2x )2 – 3 ( 2x ) – 5

= 2 ( 4x2 ) – 6x – 5

= 8x2 – 6x – 5

f(3x) = 2 ( 3x )2 –3(3x) – 5

= 2 ( 9x2 ) – 9x – 5

= 18x2 – 9x – 5

Now f (3x) = f (2x) + 81

That is, 18x2 – 9x – 5 = 8x2 – 6x – 5 + 81

18x2 – 9x –5 – 8x2 + 6x – 5 + 81 = 0

10x2 – 3x – 81 = 0

[ 10 * – 81 = – 810 ; – 30 * 27 = – 810 and – 30 + 27 = – 3 ]

10x2 – 3x – 81 = 0

10x2 – 30x + 27x – 81 = 0

10x ( x – 3 ) + 27 ( x – 3 ) = 0

( x – 3 ) ( 10x + 27 ) = 0

x – 3 = 0 or 10x + 27 = 0

x = 3 or 10x = – 27

f(3x) will be 81 more than f (2x) if x = 3 or – 2.7

12. f (x) = ( 5x – 4 )2 –3 ( 2x – 5 )2

=[ ( 5x )2 –2 * 5x * 4 + (4)2 ] – 3 [ ( 2x )2 – 2 * 2x * 5 + (5)2]

= ( 25x2 – 40x + 16 ) – 3 ( 4x2 – 20x + 25 )

= 25x2 – 40x + 16 – 12x2 + 60x – 75

= 13x2 + 20x – 59

Again g(x) = ( 4x – 3 )2 –2( x – 3 )2 + 10

= [ ( 4x )2 – 2 * 4x * 3 + (3)2] – 2[( x )2 – 2 * x * 3 + 32] + 10

= ( 16x2 – 24x + 9 ) – 2 ( x2 – 6x + 9 ) +10

= 16x2 – 24x + 9 – 2x2 + 12x –18 + 10

= 14x2 – 12x +1

f(x) =g(x)

That is 13x2 +20x – 59 = 14x2 – 12x + 1

13x2 + 20x – 59 –14x2 + 12x – 1 =0

– x2 + 32x – 60 = 0 (or) x2 – 32x + 60 =0

[ 60 = ( – 30 ) * ( – 2 ) and – 30 + ( – 2 ) = – 32 ]

x2 – 32x + 60 = 0 = x2 – 30x – 2x + 60

x ( x – 30 ) – 2 ( x – 30 ) = 0

( x – 30 ) ( x – 2 ) = 0

x – 30 = 0 or x – 2 = 0

x = 30 or x = 2

f (x) = g(x) if x = 30 or 2.