How do you use square roots to solve $2 {(x + 3)}^{2} = 8$ $2(x+3)^2=8$ ?

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How do you use square roots to solve $2 {(x + 3)}^{2} = 8$ $2(x+3)^2=8$ ?

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Answer 1 · 2018-04-15T23:04:54

$x = - 1, - 5$ $x=-1,-5$

Explanation:

Our first step would be to divide both sides by $2$ $2$ . We get:

${(x + 3)}^{2} = 4$ $(x+3)^2=4$

We can take the $\pm$ $+-$ square root of both sides, and we will now have two equations:

The first:
$x + 3 = 2$ $x+3=2$

$\Rightarrow x = - 1$ $color(blue)(=>x=-1)$

The second:
$x + 3 = - 2$ $x+3=-2$

$\Rightarrow x = - 5$ $color(blue)(=>x=-5)$

We have two equations because we need to take the $\pm$ $+-$ square root of both sides, because when you take the sqrt of a number, there will always be a positive and a negative answer.

Hope this helps!

Answer 2 · 2018-04-15T23:05:54

Express the answer with surds, it's not possible here as the surd is a perfect square.

Explanation:

So

$2 {(x + 3)}^{2} = 8$ $2(x+3)^2 = 8$

${(x + 3)}^{2} = 4$ $(x+3)^2 = 4$

$(x + 3) = \pm \sqrt{4}$ $(x+3) = +- sqrt4$
$x = \pm 2 - 3$ $x = +- 2 - 3$

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How do you use square roots to solve $2 {(x + 3)}^{2} = 8$ $2(x+3)^2=8$ ?

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