Power functions > The quadratic formula

## Theory

Take a look at the applet: Sine Functions

The general form of the quadratic formula is $f\left(x\right)=a{x}^{2}+bx+c$.
It is not immediately apparent how this function rule could be derived by transformation of the basic power function $y={x}^{2}$. That makes it difficult to find the vertex and x-intersects of the corresponding parabola.

Using a method called completing the square allows you to convert the function $f$ to the form: $f\left(x\right)=a{\left(x-p\right)}^{2}+q$ where $\left(p,q\right)$ are the coordinated of the vertex of the graph.
To do this conversion you use the following property:

${x}^{2}+2kx={\left(x+k\right)}^{2}-{k}^{2}$

Use the applet to check that $f\left(x\right)=2{x}^{2}-4x$ is the same function as $g\left(x\right)=2{\left(x-1\right)}^{2}-2$.

It is obviously very useful if by completing the square you can convert $f\left(x\right)=a{x}^{2}+bx+c$ to a form that allows you to immediately see the vertex and the axis of symmetry...

A long time ago, mathematicians derived the so-called quadratic formula.
This formula allows you to solve $a{x}^{2}+bx+c=0$ and thereby find the zeros of the quadratic equation. The general solution is:
$x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}\vee x=\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$

Below you see a proof of the quadratic formula. This means you can show that the formula is always valid. To do so you need to solve    $a{x}^{2}+bx+c=0$ in general terms by completing the square.

Assume that $a\ne 0$ (otherwise it would not be a quadratic equation!). You now divide by $a$ on both sides. This gives you:

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

Completing the square results in:

${\left(x+\frac{b}{2a}\right)}^{2}-{\left(\frac{b}{2a}\right)}^{2}+\frac{c}{a}=0$ and ${\left(x+\frac{b}{2a}\right)}^{2}={\left(\frac{b}{2a}\right)}^{2}-\frac{c}{a}=\frac{{b}^{2}-4ac}{4{a}^{2}}$

Taking the square root:

$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$

And now a few rearrangements:

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

The quadratic formula has been derived..

The expression $D={b}^{2}-4ac$ in the root is called the discriminant of the quadratic equation. Since only the root of a positive number or $0$ is itself a real number, it is the value of the discriminant that determines the number of solutions of the equation:

• $D>0$ and there are two solutions;

• $D=0$ and there is one solution (or the same solution twice);

• $D<0$ and there are no real solutions;