# How do you solve the quadratic equation by completing the square: x^2 - 6x = 16?

Mar 22, 2018

Solution: $x = 8 \mathmr{and} x = - 2$

#### Explanation:

${x}^{2} - 6 x = 16 \mathmr{and} {x}^{2} - 6 x + 9 = 16 + 9$

$\therefore {\left(x - 3\right)}^{2} = 25 \mathmr{and} \left(x - 3\right) = \pm 5$

$\therefore x = 3 + 5 \mathmr{and} x = 3 - 5$ or

$\therefore x = 8 \mathmr{and} x = - 2$

Solution: $x = 8 \mathmr{and} x = - 2$ [Ans]

Mar 22, 2018

$x = 8 \mathmr{and} x = - 2$

#### Explanation:

We have ,

${x}^{2} - 6 x = 16$

i.e. ${x}^{2} - 6 x + \left(-\right) = 16 + \left(-\right)$

To find third term, we use

$\textcolor{red}{{3}^{r d} t e r m = {\left({2}^{n d} t e r m\right)}^{2} / \left(4 \times {1}^{s t} t e r m\right)} = {\left(- 6 x\right)}^{2} / \left(4 \times {x}^{2}\right) = \frac{36 {x}^{2}}{4 {x}^{2}} = 9$

Adding both sides 9, we get

${x}^{2} - 6 x + 9 = 16 + 9 = 25$

$\implies {\left(x - 3\right)}^{2} = {\left(5\right)}^{2}$

Taking squareroot of both sides,

$\implies x - 3 = \pm 5$

$\therefore x - 3 = 5 \mathmr{and} x - 3 = - 5$

$\therefore x = 8 \mathmr{and} x = - 2$