How do you solve $\sqrt{x + 16} = \sqrt{x} + 4$ $sqrt(x+16)=sqrtx+4$ ?

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How do you solve $\sqrt{x + 16} = \sqrt{x} + 4$ $sqrt(x+16)=sqrtx+4$ ?

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Answer 1 · 2018-03-26T12:26:51

I got $x = 0$ $x=0$

Explanation:

We can square both sides:
${(\sqrt{x + 16})}^{2} = {(\sqrt{x} + 4)}^{2}$ $(sqrt(x+16))^2=(sqrt(x)+4)^2$
$x + 16 = x + 8 \sqrt{x} + 16$ $x+16=x+8sqrt(x)+16$
rearrange:
$cancel(x)cancel(+16)=cancel(x)+8sqrt(x)cancel(+16)$
so:
$8sqrt(x)=0$
and:
$x=0$

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How do you solve $\sqrt{x + 16} = \sqrt{x} + 4$ $sqrt(x+16)=sqrtx+4$ ?

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How do you solve $\sqrt{x + 16} = \sqrt{x} + 4$ $sqrt(x+16)=sqrtx+4$ ?