Solve sqrt{2x^2-9}-x=0

$$\sqrt { 2 x ^ { 2 } - 9 } - x = 0$$

Solve for x

$$x=3$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{2x^{2}-9}\right)^{2}=x^{2}$$

Calculate \sqrt{2x^{2}-9} to the power of 2 and get 2x^{2}-9.

$$2x^{2}-9=x^{2}$$

Subtract x^{2} from both sides.

$$2x^{2}-9-x^{2}=0$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-9=0$$

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)=0$$

To find equation solutions, solve x-3=0 and x+3=0.

$$x=3 x=-3$$

Substitute 3 for x in the equation \sqrt{2x^{2}-9}-x=0.

$$\sqrt{2\times 3^{2}-9}-3=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute -3 for x in the equation \sqrt{2x^{2}-9}-x=0.

$$\sqrt{2\left(-3\right)^{2}-9}-\left(-3\right)=0$$

Simplify. The value x=-3 does not satisfy the equation.

$$6=0$$

Equation \sqrt{2x^{2}-9}=x has a unique solution.

$$x=3$$