How can you solve any quadratic equation?

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Answer 1 · 2016-05-23T20:34:39

The most general methods are the quadratic formula and completing the square.

The most general methods which will cope with any quadratic equation in one variable are:

These methods are both capable of finding the roots regardless of whether they are integers, rational, irrational or even non-Real Complex numbers.

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Quadratic formula

The roots of $ax^2+bx+c = 0$ are given by the formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

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Completing the square

Given:

$ax^2+bx+c = 0$

Note that:

$a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)$

So our equation can be rewritten:

$a(x+b/(2a))^2 = b^2/(4a)-c$

Dividing both sides by $a$ we find:

$(x+b/(2a))^2 = b^2/(4a^2)-c/a$

Hence:

$x+b/(2a) = +-sqrt(b^2/(4a^2)-c/a)$

Which can be simplified to:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

Hmmm... that looks familiar.

So completing the square and the quadratic formula are kind of the same thing, but in particular cases completing the square can be a little cleaner:

For example, factoring $x^2+4x-21$ by completing the square:

$x^2+4x-21$

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

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

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

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

Or using the quadratic formula:

$x^2+4x-21$ is $ax^2+bx+c$ with $a=1$ , $b=4$ , $c=-21$

hence has zeros:

$x = (-4+-sqrt(4^2-(4*1*(-21))))/(2*1)$

$=(-4+-sqrt(16+84))/2$

$=(-4+-sqrt(100))/2$

$=(-4+-10)/2$

$=-2+-5$

i.e. $x=-7$ and $x=3$

Hence factors:

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