How do you solve $2+sqrt[x]=sqrt[x+4]$ and find any extraneous solutions?

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Answer 1 · 2016-07-04T00:50:10

$(2 + sqrt(x))^2 = (sqrt(x + 4))^2$

$4 + 4sqrtx + x = x + 4$

$4sqrt(x) = x - x + 4 - 4$

$4sqrt(x) = 0$

$(4sqrt(x))^2 = 0^2$

$16x = 0$

$x = 0$

Checking in the original equation we find this solution works. The solution set is ${0}$ .

Hopefully this helps!

How do you solve 2+sqrt[x]=sqrt[x+4]2+sqrt[x]=sqrt[x+4] and find any extraneous solutions?