How do you solve $\sqrt{x + 7} = x + 1$ $sqrt(x+7) = x + 1$ and find any extraneous solutions?

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Answer 1 · 2018-04-19T23:00:37

I tried this:

Let us square both sides:

${(\sqrt{x + 7})}^{2} = {(x + 1)}^{2}$ $(sqrt(x+7))^2=(x+1)^2$

$x + 7 = x^{2} + 2 x + 1$ $x+7=x^2+2x+1$

rearrange:

$x^{2} + x - 6 = 0$ $x^2+x-6=0$

$x_{1, 2} = \frac{- 1 \pm \sqrt{1 + 24}}{2} = \frac{- 1 \pm 5}{2}$ $x_(1,2)=(-1+-sqrt(1+24))/2=(-1+-5)/2$

giving:

$x_{1} = - 3$ $x_1=-3$ that doesn't work;
$x_{2} = 2$ $x_2=2$ .

How do you solve √x+7=x+1sqrt(x+7) = x + 1 and find any extraneous solutions?