Completing the square

Functions: Quadratic functions

Completing the square

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

Solve $x$ in $\left(x+3\right)^2={{9}\over{25}}$ .

$x=-3\pm {{3}\over{5}}$

This can be seen by

taking the root at both sides of the $=$ sign: $x+3=\pm {{3}\over{5}}$ ;
continue solving the equation by subtracting $3$ at both sides of the $=$ sign: $x=-3\pm {{3}\over{5}}$ .

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

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

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

It is rewritten in the following manner:

$\begin{array}{rcl}ax^2+bx+c &=& a\left(x^2+\frac{b}{a}x\right)+c \\ &=&a\cdot\left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c \\ &=&a\cdot \left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}\right) \\ &=&a\cdot \left(\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}+\frac{c}{a}\right) \\ &=&a\cdot \left(\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}+\frac{4ac}{4a^2}\right) \\ &=&a\cdot \left(\left(x+\frac{b}{2a}\right)^2 - \frac{b^2-4ac}{4a^2}\right) \\ \end{array}$

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

The extreme point of a quadratic function

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, $\rv{- \frac{b}{2a},-\frac{b^2-4ac}{4a}}$ 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 $x =- \dfrac{b}{2a}$ , which is $-\dfrac{b^2-4ac}{4a}$ .

In the extreme point the first term of the result of completing the square is equal to $0$ . That is the case if and only if $x+\frac{b}{2a}=0$ , meaning: if $x=-\frac{b}{2a}$ . This explains why the extreme point has $x$ coordinate $x=-\frac{b}{2a}$ .
If $a\gt0$ , then $a\cdot\left(x+\frac{b}{2a}\right)^2$ is always greater than or equal to $0$ and we are dealing with a minimum.
If $a\gt0$ , then $a\cdot\left(x+\frac{b}{2a}\right)^2$ is always less than or equal to $0$ and we are dealing with a maximum.
The corresponding value for $y$ is equal to the second term of the result of completing the square: $y=-\frac{b^2-4ac}{4a}$ .

Solve $x$ by completing the square in the equation

$x^2+2 x-1=0$

Give your answer in the form $x=x_1\lor x=x_2$ , where $x_1$ and $x_2$ are exact solutions.

$x=\sqrt{2}-1\lor x=-\sqrt{2}-1$

$\begin{array}{rcl} x^2+2 x-1&=&0\\ &&\phantom{xxx}\blue{\text{original equation}}\\ (x+1)^2-1^2-1&=&0\\ &&\phantom{xxx}\blue{\text{ }x^2+2x\text{ completed to a square}}\\ (x+1)^2&=&1^2+1\\ &&\phantom{xxx}\blue{\text{moved everything outside of the brackets to the right-hand side}}\\ (x+1)^2&=&2\\ &&\phantom{xxx}\blue{\text{simplified the right-hand side}}\\ x+1=\sqrt{2} &\lor&x+1=-\sqrt{2}\\ &&\phantom{xxx}\blue{\text{the root taken at both sides}}\\ x=\sqrt{2}-1&\lor& x=-\sqrt{2}-1\\ &&\phantom{xxx}\blue{1 \text{ subtracted from both sides}}\\ \end{array}$

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