How do you solve $\sqrt{x + 2} = x - 4$ $sqrt(x+2) = x-4$ ?

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Answer 1 · 2016-05-10T23:01:49

$x = 7$ $x=7$

First square both sides of the equation, noting that this may introduce spurious solutions:

$x + 2 = {(x - 4)}^{2} = x^{2} - 8 x + 16$ $x+2 = (x-4)^2 = x^2-8x+16$

Subtract $x + 2$ $x+2$ from both ends to get:

$0 = x^{2} - 9 x + 14 = (x - 2) (x - 7)$ $0 = x^2-9x+14 = (x-2)(x-7)$

So $x = 2$ $x=2$ or $x = 7$ $x=7$

If $x = 2$ $x=2$ then $\sqrt{x + 2} = \sqrt{4} = 2 \neq - 2 = 2 - 4 = x - 4$ $sqrt(x+2) = sqrt(4) = 2 != -2 = 2 - 4 = x-4$

If $x = 7$ $x=7$ then $\sqrt{x + 2} = \sqrt{9} = 3 = 7 - 4 = x - 4$ $sqrt(x+2) = sqrt(9) = 3 = 7-4 = x-4$

So $x = 7$ $x=7$ is the only valid solution of the original equation.

How do you solve √x+2=x−4sqrt(x+2) = x-4?