How do you solve using the completing the square method x^2 + 6x = 7?

Jul 20, 2017

$x = 1 \mathmr{and} - 7$

Explanation:

${x}^{2} + 6 x = 7$

half the coefficient of $x$ and square it

$6 \rightarrow {3}^{2}$

$\textcolor{red}{{x}^{2} + 6 x + {3}^{2}} - {3}^{2} = 7$

$\textcolor{w h i t e}{= =} \uparrow$ the terms in red form a perfect square

$\textcolor{red}{\left({x}^{2} + 6 x + {3}^{2}\right)} - 9 = 7$

$\textcolor{w h i t e}{= =} \downarrow$

$\implies \textcolor{red}{{\left(x + 3\right)}^{2}} = 7 + 9$

${\left(x + 3\right)}^{2} = 16$

now solve for $x$

$\implies x + 3 = \pm \sqrt{16}$

$x + 3 = \pm 4$

$x = - 3 \pm 4$

$x = 1 \mathmr{and} - 7$