# How do you solve f(x)=2x^2-12x+17 by completing the square?

Apr 23, 2017

#### Answer:

$3 \pm \sqrt{\frac{1}{2}}$

#### Explanation:

$f \left(x\right) = 2 {x}^{2} - 12 x + 17 \implies 2 \left({x}^{2} - 6 x\right) + 17$

so,

$2 {\left(x - 3\right)}^{2} - 18 + 17$

Hence,

$2 {x}^{2} - 12 x + 17 = 2 {\left(x - 3\right)}^{2} - 1$

By 'solving' I assume you mean $2 {x}^{2} - 12 x + 17 = 0$

i.e. $2 {\left(x - 3\right)}^{2} - 1 = 0$

$\implies {\left(x - 3\right)}^{2} = \frac{1}{2}$

=> x - 3 = ± sqrt(1/2)

so, x = 3 ± sqrt(1/2)

:)>

Apr 23, 2017

#### Answer:

$x = 3 \pm \frac{\sqrt{2}}{2}$

#### Explanation:

To $\textcolor{b l u e}{\text{complete the square}}$

add ${\left(\frac{1}{2} \text{ coefficient of x-term}\right)}^{2}$

Require coefficient of ${x}^{2}$ term to be 1

$f \left(x\right) = 2 \left({x}^{2} - 6 x\right) + 17$

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({x}^{2} - 6 x \textcolor{red}{+ 9 - 9}\right) + 17$

Since we have added +9 which is not there we must also subtract 9

$f \left(x\right) = 2 {\left(x - 3\right)}^{2} - 18 + 17$

$\Rightarrow f \left(x\right) = 2 {\left(x - 3\right)}^{2} - 1$

To solve $\textcolor{b l u e}{\text{equate f(x) to zero}}$

$\Rightarrow 2 {\left(x - 3\right)}^{2} - 1 = 0$

$\Rightarrow 2 {\left(x - 3\right)}^{2} = 1$

$\Rightarrow {\left(x - 3\right)}^{2} = \frac{1}{2}$

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

$\sqrt{{\left(x - 3\right)}^{2}} = \pm \sqrt{\frac{1}{2}}$

$\Rightarrow x - 3 = \pm \frac{1}{\sqrt{2}}$

$\Rightarrow x = 3 \pm \frac{1}{\sqrt{2}} = 3 \pm \frac{\sqrt{2}}{2} \leftarrow \textcolor{red}{\text{ rationalise denominator}}$