How can you solve any quadratic equation?

Question

Quincy Doyle · Accepted Answer

The most general methods which will cope with any quadratic equation in one variable are:
The quadratic formula.
Completing the square.
Whether the roots are integers, rational, irrational, or even non-Real Complex numbers, both methods can both locate them.
Quadratic formula
The roots of

a x^{2} + b x + c = 0

are given by the formula:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Completing the square
Given:

a x^{2} + b x + c = 0

Note that:

a {(x + \frac{b}{2 a})}^{2} = a x^{2} + b x + \frac{b^{2}}{4 a}

We can now rewrite our equation as follows:

a {(x + \frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a} - c

Dividing both sides by a we find:

{(x + \frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} - \frac{c}{a}

Thus:

x + \frac{b}{2 a} = \pm \sqrt{\frac{b^{2}}{4 a^{2}} - \frac{c}{a}}

Which can be simplified to:

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

So the quadratic formula and completing the square are somewhat equivalent, however in some situations completing the square can be a little cleaner:
For example, factoring

x^{2} + 4 x - 21

by completing the square:

x^{2} + 4 x - 21

= x^{2} + 4 x + 4 - 25

= {(x + 2)}^{2} - 5^{2}

= ((x + 2) - 5) ((x + 2) + 5)

= (x - 3) (x + 7)

You may also use the quadratic formula:

x^{2} + 4 x - 21

is

a x^{2} + b x + c

with

a = 1

,

b = 4

,

c = - 21

thus has zeros:

x = \frac{- 4 \pm \sqrt{4^{2} - (4 \cdot 1 \cdot (- 21))}}{2 \cdot 1}

= \frac{- 4 \pm \sqrt{16 + 84}}{2}

= \frac{- 4 \pm \sqrt{100}}{2}

= \frac{- 4 \pm 10}{2}

= - 2 \pm 5

i.e.

x = - 7

and

x = 3

Therefore factors:

(x + 7) (x - 3)

How can you solve any quadratic equation?

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