What is the quadratic formula?

Question

1675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T19:13:35

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

Given a quadratic equation in the standard form:

$ax^2+bx+c = 0$

the roots are given by the quadratic formula:

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

This is a very useful formula to memorise, but where does it come from?

Given:

$ax^2+bx+c = 0$

Note that:

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

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

So:

$0 = ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))$

Add $(b^2/(4a)-c)$ to both ends and transpose to get:

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

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

Divide both sides by $a$ to get:

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

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

Take the square root of both sides, allowing for both positive and negative square roots:

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

Subtract $b/(2a)$ from both sides to get:

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

Note that the quadratic formula will always work, but it will only give real values if $b^2-4ac >= 0$ , resulting in real square roots.