Solve sqrt{2-3x^2}=x

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

Solve for x

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{2-3x^{2}} to the power of 2 and get 2-3x^{2}.

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

Subtract x^{2} from both sides.

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

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$2-4x^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}=-2$$

Divide both sides by -4.

$$x^{2}=\frac{-2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out -2.

$$x^{2}=\frac{1}{2}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{2}}{2} x=-\frac{\sqrt{2}}{2}$$

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

$$\sqrt{2-3\times \left(\frac{\sqrt{2}}{2}\right)^{2}}=\frac{\sqrt{2}}{2}$$

Simplify. The value x=\frac{\sqrt{2}}{2} satisfies the equation.

$$\frac{1}{2}\times 2^{\frac{1}{2}}=\frac{1}{2}\times 2^{\frac{1}{2}}$$

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

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

Simplify. The value x=-\frac{\sqrt{2}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{2}\times 2^{\frac{1}{2}}=-\frac{1}{2}\times 2^{\frac{1}{2}}$$

Equation \sqrt{2-3x^{2}}=x has a unique solution.

$$x=\frac{\sqrt{2}}{2}$$