How do you solve x^2+10x+16=0 by completing the squares?

Jul 17, 2018

$x = - 8 , - 2$

Explanation:

Given: ${x}^{2} + 10 x + 16 = 0$

To complete the square put the two $x$ terms on the left and the constant on the right of the equation:

$\textcolor{red}{{x}^{2} + 10 x} = - 16$

Complete the square by multiplying the $x$-term by $\frac{1}{2}$:
$\frac{1}{2} \cdot 10 = 5$

and adding the square of this number to the right side of the equation: ${5}^{2} = 25$

${\left(x + 5\right)}^{2} = - 16 \textcolor{b l u e}{+ 25}$

UNDERSTANDING CHECK:
${\left(x + 5\right)}^{2} = \textcolor{red}{{x}^{2} + 10 x} \textcolor{b l u e}{+ 25} .$

The $+ 25$ was not in the original equation. If we add $+ 25$ to one side of the equation, we must add the same amount to the other side of the equation to keep it balanced.

${\left(x + 5\right)}^{2} = 9$

To solve, square root both sides of the equation:

$\sqrt{{\left(x + 5\right)}^{2}} = \pm \sqrt{9}$

$x + 5 = \pm 3$

$x = - 5 \pm 3$

$x = - 5 + 3 = - 2 , \text{ } x = - 5 - 3 = - 8$

$x = - 8 , - 2$

Jul 17, 2018

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

Explanation:

The process of completing the square is done by adding a missing term to an expression so as to create the square of a binomial.

In ${x}^{2} + 10 x + 16 = 0 , \text{ } 16$ is not the required constant, so move it to the right side.

${x}^{2} + 10 x \text{ } = - 16$

The required constant is determined from $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2}}$ where $b = 10$

$\textcolor{b l u e}{{\left(\frac{10}{2}\right)}^{2} = {5}^{2} = 25}$

Add this to both sides of the equation:

${x}^{2} + 10 x \textcolor{b l u e}{+ 25} = - 16 \textcolor{b l u e}{+ 25}$

The left side is now the square of a binomial, ie a perfect square.

$\text{ "(x+5)^2 = 9" } \leftarrow$ find the square root of both sides.

$\text{ } x + 5 = \pm \sqrt{9} = \pm 3$

$\text{ } x = \pm 3 - 5$

$x = + 3 - 5 = - 2 \text{ "or" } x = - 3 - 5 = - 8$