How do you solve $x=sqrt( -7x+8)$ and find any extraneous solutions?

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Answer 1 · 2016-07-16T04:05:44

$(x)^2 = (sqrt(-7x + 8))^2$

$x^2 = -7x + 8$

$x^2 + 7x - 8 = 0$

$(x + 8)(x - 1) = 0$

$x = -8 and 1$

Checking in the original équation, we find only $x = 1$ works. Hence, ${1}$ is the solution set.

Hopefully this helps!

How do you solve x=sqrt( -7x+8)x=sqrt( -7x+8) and find any extraneous solutions?