Solving Quadratic Equations

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.

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

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

Answers to questions

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

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Friday, May 9, 2014