# How do you answer this?

Jan 31, 2018

$x = 4 + \sqrt{19}$ and $x = 4 - \sqrt{19}$

#### Explanation:

Using complete the square:

Half term of $x$

$\frac{8}{2} = 4 x$

Put inside bracket and square it, adding on the constant.

${\left(x - 4\right)}^{2} - 3$

Take away the squared number of the term of $x$ in the bracket:

${\left(x - 4\right)}^{2} - 16 - 3$

Collect like terms

$- 16 - 3 = - 19$

${\left(x - 4\right)}^{2} - 19$

This is the completed the square format, but now we solve.

${\left(x - 4\right)}^{2} - 19 = 0$

Add $19$

${\left(x - 4\right)}^{2} = 19$

Get rid of the squared bracket, by square rooting $19$.

Remember the $\pm$

$x - 4 = \pm \sqrt{19}$

$x = 4 \pm \sqrt{19}$

This means the answer will be either $x = 4 + \sqrt{19}$ and $x = 4 - \sqrt{19}$ as we usually have $2$ solutions.

Feb 6, 2018

$x = 4 \pm \sqrt{19}$

#### Explanation:

$\text{solve using the method of "color(blue)"completing the square}$

${x}^{2} - 8 x - 3 = 0$

• " the coefficient of the "x^2" term must be 1 which it is"

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 8 x$

${x}^{2} + 2 \left(- 4\right) x \textcolor{red}{+ 16} \textcolor{red}{- 16} - 3 = 0$

$\Rightarrow {\left(x - 4\right)}^{2} - 19 = 0$

$\Rightarrow {\left(x - 4\right)}^{2} = 19$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x - 4\right)}^{2}} = \pm \sqrt{19} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x - 4 = \pm \sqrt{19}$

$\text{add 4 to both sides}$

$\Rightarrow x = 4 \pm \sqrt{19} \leftarrow \textcolor{red}{\text{exact values}}$