### 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+1\right)^2={{9}\over{4}}#.

This can be seen by

- taking the root at both sides of the #=# sign: #x+1=\pm {{3}\over{2}}#;
- continue solving the equation by subtracting #1# at both sides of the #=# sign: #x=-1\pm {{3}\over{2}}#.

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}#.

\[x^2+6 x-8=0\]

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

#\begin{array}{rcl}

x^2+6 x-8&=&0\\

&&\phantom{xxx}\blue{\text{original equation}}\\

(x+3)^2-3^2-8&=&0\\

&&\phantom{xxx}\blue{\text{ }x^2+6x\text{ completed to a square}}\\

(x+3)^2&=&3^2+8\\

&&\phantom{xxx}\blue{\text{everything outside of the brackets moved to the right hand side}}\\

(x+3)^2&=&17\\

&&\phantom{xxx}\blue{\text{simplified the right hand side}}\\

x+3=\sqrt{17} &\lor&x+3=-\sqrt{17}\\

&&\phantom{xxx}\blue{\text{the root taken at both sides}}\\

x=\sqrt{17}-3&\lor& x=-\sqrt{17}-3\\

&&\phantom{xxx}\blue{3 \text{ subtracted from both sides}}\\

\end{array}

#

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