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

Apr 18, 2018

$x = 0 , 2 , - \frac{3 \sqrt{2}}{2} \le x \le \frac{3 \sqrt{2}}{2}$

#### Explanation:

Restrictions: $- 2 {x}^{2} + 9 \ge 0$

$\implies - \frac{3 \sqrt{2}}{2} \le x \le \frac{3 \sqrt{2}}{2}$

$x + \sqrt{- 2 {x}^{2} + 9} = 3$

$\sqrt{- 2 {x}^{2} + 9} = 3 - x$ subtract $x$ from both sides

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

$3 {x}^{2} - 6 x = 0$

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

$x = 0 , x = 2$