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.
The following example illustrates the method.
x = 2,  3/2 are the roots of the equation.
Determine whether this equation reduces to the quadratic form.
( x + a ) ( x + b ) = ab/p>
6x^{2} – 13x + 6 = 0
6x^{2} – 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.
x^{2} – 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 + ( a^{2} – b^{2} ) = 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
x^{2} + x ( c – b ) + ( c – a ) ( a – b ) = 0
[ c – a + a – b = c – b ]
Therefore x^{2} + 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 ).

( 2x + 1 ) /( a + 1 ) + 2a /x = 5
{x ( 2x + 1 ) + 2a ( a + 1 )} / x ( a + 1 ) = 5
2x^{2} + x + 2a^{2} + 2a = 5x ( a + 1 )
2x^{2} + x + 2a^{2} + 2a = 5ax + 5x
2x^{2} + x + 2a^{2} + 2a – 5ax – 5x = 0
2x^{2} – 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.

( a – x )^{2} + ( b – x )^{2} = ( a – b )^{2}
a^{2} – 2ax + x^{2} + b^{2} – 2bx + x^{2} = a^{2} – 2ab + b^{2}
2x^{2} – 2ax – 2bx + 2ab = 0
2x^{2} – 2x ( a + b ) + 2ab = 0
[ 2 *2ab = 4ab; 4ab = 2a * 2b and (– 2a) + (– 2b) = – ( 2a + 2b )]
2x^{2} – 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.
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 + a^{2} = – bx
bx + ax + bx = – ab – a2
2bx + ax = – ab – a^{2}
( 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 )
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 – b^{2}
Therefore x = b + ab – b^{2} / a – b = b ( a – b + 1 )/ ( a – b )
Therefore x = a or b (a – b + 1) / ( a – b ).
a /( x – b ) + b/( x – a ) = 2
( ax – a^{2} + bx – b^{2} ) / { ( x – b ) ( x – a ) } = 2
ax – a^{2} + bx – b^{2} = 2( x –b ) ( x – a )
ax – a^{2} + bx – b^{2} = 2( x^{2} – ax – bx + ab )
ax – a^{2} + bx – b^{2} = 2x^{2} – 2ax – 2bx + 2ab
2x^{2} – 2ax – 2bx + 2ab – ax – b x + a^{2} + b^{2} = 0
2x^{2} – 3x ( a + b ) + ( a^{2} + 2ab + b^{2} ) = 0
2x^{2} – 3x ( a + b ) + ( a + b )2 = 0
2x^{2} – 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.
( x^{2} + ax + a^{2} ) / ( x^{2} – ax + a^{2} ) = 13 / 7
7x2 + 7ax + 7a2 = 13x2 – 13ax + 13a^{2}
13x^{2} – 13ax + 13a^{2} – 7x2 – 7ax – 7a^{2} =0
6x^{2} – 20ax + 6a^{2} = 0
3x^{2} – 10ax + 3a^{2} = 0
[ 3 * 3a^{2} = 9a^{2} ; 9a * a and ( – 9a ) + ( – a ) = – 10a ]
3x^{2} – 10ax + 3a^{2} = 0 = 3x^{2} – 9ax – ax + 3a^{2}
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.
3 f(x) = 3 [ 3x2 – 2x + 1]
= 9x^{2} – 6x + 3
4g (x) = 4 [ 2x – 3/2] = 8x^{2} – 6
3 f(x) = 4g (x)
9x^{2} – 6x + 3 = 8x^{2} – 6
9x^{2} – 6x + 3 – 8x^{2} + 6 = 0
x^{2} – 6x + 9 = 0
( x – 3 )2 = 0 or ( x – 3 ) =0
x = 3
if x = 3, then 3 f(x) = 4g (x)
f(x) = 2x^{2} – 3x – 5
f(2x) = 2 ( 2x )2 – 3 ( 2x ) – 5
= 2 ( 4x^{2} ) – 6x – 5
= 8x^{2} – 6x – 5
f(3x) = 2 ( 3x )2 –3(3x) – 5
= 2 ( 9x^{2} ) – 9x – 5
= 18x^{2} – 9x – 5
Now f (3x) = f (2x) + 81
That is, 18x^{2} – 9x – 5 = 8x^{2} – 6x – 5 + 81
18x^{2} – 9x –5 – 8x^{2} + 6x – 5 + 81 = 0
10x2 – 3x – 81 = 0
[ 10 * – 81 = – 810 ; – 30 * 27 = – 810 and – 30 + 27 = – 3 ]
10x^{2} – 3x – 81 = 0
10x^{2} – 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
f (x) = ( 5x – 4 )2 –3 ( 2x – 5 )2
=[ ( 5x )^{2} –2 * 5x * 4 + (4)^{2} ] – 3 [ ( 2x )^{2} – 2 * 2x * 5 + (5)^{2}]
= ( 25x^{2} – 40x + 16 ) – 3 ( 4x^{2} – 20x + 25 )
= 25x^{2} – 40x + 16 – 12x^{2} + 60x – 75
= 13x^{2} + 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
= ( 16x^{2} – 24x + 9 ) – 2 ( x^{2} – 6x + 9 ) +10
= 16x^{2} – 24x + 9 – 2x^{2} + 12x –18 + 10
= 14x^{2} – 12x +1
f(x) =g(x)
That is 13x^{2} +20x – 59 = 14x^{2} – 12x + 1
13x^{2} + 20x – 59 –14x^{2} + 12x – 1 =0
– x^{2} + 32x – 60 = 0 (or) x^{2} – 32x + 60 =0
[ 60 = ( – 30 ) * ( – 2 ) and – 30 + ( – 2 ) = – 32 ]
x^{2} – 32x + 60 = 0 = x^{2} – 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.