# Quadratic function – Completing the square

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A quadratic function $$f$$ is given by $$f(x)=ax^2+bx+c$$ with $$a$$, $$b$$ and $$c$$ being real numbers and $$a\neq 0$$.
The graph associated with such a function is called a parabola.
When $$a\gt0$$, we speak of a parabola opening upwards, and when $$a\lt0$$, we speak of a parabola opening downwards.

The significance of these names becomes evident from the graph below of the parabola opening upwards of $$x^2+8$$ and the parabola opening downwards of $$-x^2-8$$.

To find the zeros of a quadratic function, we need do find the points of intersection of the parabola with the $$x$$-axis. We can do this in three ways:

• completing the square
• $$abc$$-formula
• factorization

In what follows, we will discuss the completing the square method.

An equation like $$\left(x+2\right)^2=3$$ can easily be solved by taking the root:

Solve $$x$$ in $$\left(x+4\right)^2={{16}\over{25}}$$.

Take the root at both sides of the $$=$$ sign: $$x+4=\pm {{4}\over{5}}$$;

Next, solve the equation: $$x=-4\pm {{4}\over{5}}$$;

$$x=-4\pm {{4}\over{5}}$$

However, this will not work in general for a quadratic equation like $$x^2+8x-1=0$$, because the unknown $$x$$ occurs twice. But there is a method to obtain the first form from the second one:

Completing the square is rewriting a quadratic expression in $$x$$ as an expression in which $$x$$ only occurs once, in the base of a second power. To be precise, if $$a$$, $$b$$ and $$c$$ are real numbers, then

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

With this method we cannot only solve quadratic equations, but also determine what the extreme point of a parabola is:

The quadratic polynomial $$ax^2+bx+c$$ in which $$a\ne0$$, can be written as

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

In particular, $$\left[-\frac{b}{2a},-\frac{{b}^{2}-4a·c}{4a}\right]$$ is an extreme:

• If $$a\gt0$$, then the extreme is the lowest point of the parabola opening upwards.
• If $$a\lt0$$, then the extreme is the highest point of the parabola opening downwards.

In other words, the quadratic function $$ax^2+bx+c$$ in $$x$$ has a minimum or maximum (depending on $$a\gt0$$ or $$a\lt0$$) for $$x=-\frac{b}{2a}$$, which is $$-\frac{b^2-4ac}{4a}$$.

Let’s practice!

Solve $$x$$ by completing the square in the equation

$$x^2-6 x-9=0$$

Give your answer in the form $$x=a\lor x=b$$.

To find this answer, we complete the square in the following manner:

$$\begin{array}{rcl} (x-3)^2-3^2-9&=&0\\ &&\phantom{xx}\color{blue}{\text{ }x^2-6x\text{ completed to a square}}\\ (x-3)^2&=&3^2+9\\ &&\phantom{xx}\color{blue}{\text{everything outside of the brackets moved to the right hand side}}\\ (x-3)^2&=&18\\ &&\phantom{xx}\color{blue}{\text{simplified the right hand side}}\\ x-3=\sqrt{18} &\lor&x-3=-\sqrt{18}\\ &&\phantom{xx}\color{blue}{\text{the root taken at both sides}}\\ x=3\cdot \sqrt{2}+3&\lor& x=3-3\cdot \sqrt{2}\\ &&\phantom{xx}\color{blue}{-3 \text{ subtracted from both sides}}\\ \end{array}$$

$$x=3\cdot \sqrt{2}+3\lor x=3-3\cdot \sqrt{2}$$