How do you solve $\sqrt{4} = \sqrt{x + 2}$ $sqrt4=sqrt(x+2)$ and check your solution?

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How do you solve $\sqrt{4} = \sqrt{x + 2}$ $sqrt4=sqrt(x+2)$ and check your solution?

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Answer 1 · 2017-04-20T16:08:17

$x = 2$ $x=2$

Explanation:

Raise both sides to the power of $^2$ $^2$ , doing so reduces the square roots on both sides of the equation to $1$ $1$ and makes this problem simplier without changing the question.

${(\sqrt{4})}^{2} = {(\sqrt{x + 2})}^{2}$ $(sqrt4)^2=(sqrt(x+2))^2$

$4 = x + 2$ $4=x+2$

$x = 2$ $x=2$

Checking the solution by substituiting $x = 2$ $x=2$ into the original problem,

$\sqrt{4} = \sqrt{2 + 2}$ $sqrt4=sqrt(2+2$

$\sqrt{4} = \sqrt{4}$ $sqrt4=sqrt4$

Therefore $x = 2$ $x=2$ is the solution

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How do you solve $\sqrt{4} = \sqrt{x + 2}$ $sqrt4=sqrt(x+2)$ and check your solution?

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