# Proof of the Quadratic formula

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The Quadratic Formula used to solve equations of the form $ax^2+bx+c=0 \,\!$. It is:

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

Proof:

Dividing our quadratic equation by $a\,\!$ (which is allowed because $a\,\!$ is non-zero), we have

$x^2 + \frac{b}{a} x + \frac{c}{a}=0$

which is equivalent to

$x^2 + \frac{b}{a} x= -\frac{c}{a}.$

The equation is now in a form in which we can conveniently complete the square. To "complete the square" is to add a constant (i.e., in this case, a quantity that does not depend on $x\,\!$) to the expression to the left of "$=\,\!$", that will make it a perfect square trinomial of the form $x^2+2xy+y^2\,\!$. Since $2xy\,\!$ in this case is $\frac{b}{a} x$, we must have $y = \frac{b}{2a}$, so we add the square of $\frac{b}{2a}$ to both sides, getting

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

The left side is now a perfect square; it is the square of $\left(x + \frac{b}{2a}\right)$. The right side can be written as a single fraction; the common denominator is $4a^2\,\!$. We get

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$

Taking square roots of both sides yields

$\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac\ }}{|2a|}\Leftrightarrow$$x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\ }}{2a}$

Subtracting $\frac{b}{2a}$ from both sides, we get

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