How do you solve $x+sqrt(-2x^2+9)=3$ and identify any restrictions?

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Answer 1 · 2018-04-18T04:18:25

$x=0, 2, -(3sqrt2)/2<=x<=(3sqrt2)/2$

Restrictions: $-2x^2+9>=0$

$=>-(3sqrt2)/2<=x<=(3sqrt2)/2$

$x+sqrt(-2x^2+9)=3$

$sqrt(-2x^2+9)=3-x$ subtract $x$ from both sides

$-2x^2+9=9-6x+x^2$ square both sides

$3x^2-6x=0$

$3x(x-2)=0$ solve quadratic equation

$x=0,x=2$

How do you solve x+sqrt(-2x^2+9)=3x+sqrt(-2x^2+9)=3 and identify any restrictions?