# Where does the quadratic formula come from? [duplicate]

Everywhere I look, the $ax^2+bx+c$ portion of the quadratic formula is listed as given.

Does anyone know where this comes from?

Edit

How can we prove that (x+y)^2 = ax^2+bx+c?

## marked as duplicate by Hakim, Dan Rust, lhf, mrf, Hans LundmarkJun 13 '14 at 14:04

• @DavidMitra I think he actually asks why we write the quadratic as $ax^2+bx+c$ – N. S. Jun 13 '14 at 13:49
Let $ax^2+bx+c=0$, thus, $$x^2+\frac{b}{a} x+\frac{c}{a}=0$$ $$\Longrightarrow x^2+\frac{b}{a}x=-\frac{c}{a}$$ $$\Longrightarrow x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2.$$ We can factorise the left hand side, thus $$x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left(x+\frac{b}{2a}\right)^2$$therefore $$\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2$$ $$\Longrightarrow x+\frac{b}{2a}=\pm\sqrt{-\frac{c}{a}+\left(\frac{b}{2a}\right)^2}$$ $$\Longrightarrow x=-\frac{b}{2a}\pm\sqrt{-\frac{c}{a}+\left(\frac{b}{2a}\right)^2}$$ $$=-b\pm\frac{\sqrt{-\frac{c}{a}(2a)^2+\left(\frac{b}{2a}\right)^2(2a)^2}}{2a}$$ $$=\frac{-b\pm \sqrt{-4ac+b^2}}{2a}.$$
If you are actually asking why we write quadratics in the form $$ax^2+bx+c$$ then we must note that a quadratic is a polynomial. A polynomial is of the form $$a_nx^n+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_2x^2+a_1x+a_0.$$ Thus, a linear equation $a_1x+a_0$ is a polynomial, as is a quartic, $a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$. A polynomial, as shown above, is a sum of powers of a variable $x$ that are multlied by constants $a_0,a_1,...,a_n\in \Bbb{C}.$ A quadratic is a polynomial hence we write it in the form of a stand polynomial $ax^2+bx+c$.