maths.com Algebra Solve A Quadratic Equation By Factorisation

Algebra

Solve A Quadratic Equation By Factorisation

* Introduction
* Example (1) x² - 5x + 4 = 0
* Example (2) x² - 2x - 8 = 0

* Example (1) x² + 9x = 0
* Example (2) x² - 5x = - 6
* Example (3) x² - 8 = 7x
* Example (4) x² = 2x

* Example (1) 6 - 10x = 4x²
* Example (2) x(x+ 5) = 6
* Example (3) 25x² = 49

* Example (1) (x + 2)(x+ 5) = - 2
* Example (2) 3x³ = 75x
[Note that Example (2)
is a cubic equation rather than a quadratic
but further demonstrates the use of a product equal to zero]

* Example (1) (r + 7)(r - 9) = 0
* Example (2) (13r + 7)(6r - 18) = 0
* Example (3) 4t(9t + 5) = 0

* Example (1) b² + 13b + 36 = 0
* Example (2) b² = 18 - 3b
* Example (3) v(8v + 1) = 7
* Example (4) 8v² + v - 7 = 0
* Example (5) 9t³ = 36t
[Note that Example (5)
is a cubic equation rather than a quadratic
but further demonstrates the use of a product equal to zero]

* Example (1) (x + 3)(3x - 5) = 0
* Example (2) x² + 2x - 8 = 0
* Example (3) x² + 2x = 10 - x

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Worked Example

x² - 12x - 28 = 0

Create 2 sets of brackets = 0
( )( ) = 0

Put x at the start of each set
(x )(x ) = 0

Need 2 numbers which
multiply to give -28
and add to give -12
1      -28
2      -14
4      -7
7      -4
14      -2
28      -1
+2 × -14 = -28
+2 + -14 = -12

Put one number into each set of brackets
(x + 2)(x - 14) = 0

We only get 0 by multiplication
if one of the things we start with is 0
(x + 2)(x - 14) = 0
so (x + 2) = 0 => x = -2
or (x - 14) = 0 => x = +14

Edited by Dr David Cornelius an Independent Private Maths Tutor with over 25 years of experience and The Secretary of The Association of Tutors in the UK for 15 years.