How do you solve $x+3=sqrt(18-2x^2)$ algebraically?

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Answer 1 · 2016-12-26T19:26:05

${-3}$

Square both sides.

$(x + 3)^2 = (sqrt(18 - 2x^2))^2$

$x^2 + 6x + 9 = 18 - 2x^2$

$3x^2 + 6x - 9 = 0$

$3(x^2 + 2x - 3) = 0$

$x^2 + 2x - 3 = 0$

$(x + 3)(x -1) = 0$

$x = -3 and 1$

Checking in the original equation, you will find that only $x = -3$ will work.

Hopefully this helps!

How do you solve x+3=sqrt(18-2x^2)x+3=sqrt(18-2x^2) algebraically?