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2023-02-26

How can you solve any quadratic equation?

Quincy Doyle

The most general methods which will cope with any quadratic equation in one variable are:
Completing the square.
Whether the roots are integers, rational, irrational, or even non-Real Complex numbers, both methods can both locate them.
The roots of $a{x}^{2}+bx+c=0$ are given by the formula:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Completing the square
Given:
$a{x}^{2}+bx+c=0$
Note that:
$a{\left(x+\frac{b}{2a}\right)}^{2}=a{x}^{2}+bx+\frac{{b}^{2}}{4a}$
We can now rewrite our equation as follows:
$a{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}}{4a}-c$
Dividing both sides by a we find:
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
Thus:
$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}}$
Which can be simplified to:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
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:
${x}^{2}+4x-21$
$={x}^{2}+4x+4-25$
$={\left(x+2\right)}^{2}-{5}^{2}$
$=\left(\left(x+2\right)-5\right)\left(\left(x+2\right)+5\right)$
$=\left(x-3\right)\left(x+7\right)$
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:
$x=\frac{-4±\sqrt{{4}^{2}-\left(4\cdot 1\cdot \left(-21\right)\right)}}{2\cdot 1}$
$=\frac{-4±\sqrt{16+84}}{2}$
$=\frac{-4±\sqrt{100}}{2}$
$=\frac{-4±10}{2}$
$=-2±5$
i.e. $x=-7$ and $x=3$
Therefore factors:
$\left(x+7\right)\left(x-3\right)$

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