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2023-02-26

How can you solve any quadratic equation?

Quincy Doyle

Beginner2023-02-27Added 8 answers

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 $a{x}^{2}+bx+c=0$ are given by the formula:

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

Completing the square

Given:

$a{x}^{2}+bx+c=0$

Note that:

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

We can now rewrite our equation as follows:

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

Dividing both sides by a we find:

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

Thus:

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

Which can be simplified to:

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

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 ${x}^{2}+4x-21$ by completing the square:

${x}^{2}+4x-21$

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

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

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

$=(x-3)(x+7)$

You may also use the quadratic formula:

${x}^{2}+4x-21$ is $a{x}^{2}+bx+c$ 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:

$(x+7)(x-3)$

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 $a{x}^{2}+bx+c=0$ are given by the formula:

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

Completing the square

Given:

$a{x}^{2}+bx+c=0$

Note that:

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

We can now rewrite our equation as follows:

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

Dividing both sides by a we find:

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

Thus:

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

Which can be simplified to:

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

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 ${x}^{2}+4x-21$ by completing the square:

${x}^{2}+4x-21$

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

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

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

$=(x-3)(x+7)$

You may also use the quadratic formula:

${x}^{2}+4x-21$ is $a{x}^{2}+bx+c$ 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:

$(x+7)(x-3)$