# How do you solve using completing the square method x^2-7x+8=0?

Jul 12, 2017

Solution: $x = \frac{1}{2} \left(7 + \sqrt{17}\right) \mathmr{and} x = \frac{1}{2} \left(7 - \sqrt{17}\right)$

#### Explanation:

${x}^{2} - 7 x + 8 = 0 \mathmr{and} {x}^{2} - 7 x + {\left(\frac{7}{2}\right)}^{2} - \frac{49}{4} + 8 = 0$ or

${\left(x - \frac{7}{2}\right)}^{2} - \frac{17}{4} = 0 \mathmr{and} {\left(x - \frac{7}{2}\right)}^{2} = \frac{17}{4}$ or

$\left(x - \frac{7}{2}\right) = \pm \sqrt{\frac{17}{4}} \mathmr{and} x = \frac{7}{2} \pm \sqrt{\frac{17}{4}}$ or

$x = \frac{7}{2} \pm \frac{\sqrt{17}}{2} \mathmr{and} x = \frac{1}{2} \left(7 \pm \sqrt{17}\right)$

Solution: $x = \frac{1}{2} \left(7 + \sqrt{17}\right) \mathmr{and} x = \frac{1}{2} \left(7 - \sqrt{17}\right)$ [Ans]