Solve 1/2x^2=sqrt{8-x^2}

$$\frac { 1 } { 2 } x ^ { 2 } = \sqrt { 8 - x ^ { 2 } }$$

Solve for x

$$x=2$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\frac{1}{2}x^{2}\right)^{2}=\left(\sqrt{8-x^{2}}\right)^{2}$$

Expand \left(\frac{1}{2}x^{2}\right)^{2}.

$$\left(\frac{1}{2}\right)^{2}\left(x^{2}\right)^{2}=\left(\sqrt{8-x^{2}}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\left(\frac{1}{2}\right)^{2}x^{4}=\left(\sqrt{8-x^{2}}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}x^{4}=\left(\sqrt{8-x^{2}}\right)^{2}$$

Calculate \sqrt{8-x^{2}} to the power of 2 and get 8-x^{2}.

$$\frac{1}{4}x^{4}=8-x^{2}$$

Subtract 8 from both sides.

$$\frac{1}{4}x^{4}-8=-x^{2}$$

Add x^{2} to both sides.

$$\frac{1}{4}x^{4}-8+x^{2}=0$$

Substitute t for x^{2}.

$$\frac{1}{4}t^{2}+t-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute \frac{1}{4} for a, 1 for b, and -8 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{4}\left(-8\right)}}{\frac{1}{4}\times 2}$$

Do the calculations.

$$t=\frac{-1±3}{\frac{1}{2}}$$

Solve the equation t=\frac{-1±3}{\frac{1}{2}} when ± is plus and when ± is minus.

$$t=4 t=-8$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=2 x=-2$$

Substitute 2 for x in the equation \frac{1}{2}x^{2}=\sqrt{8-x^{2}}.

$$\frac{1}{2}\times 2^{2}=\sqrt{8-2^{2}}$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute -2 for x in the equation \frac{1}{2}x^{2}=\sqrt{8-x^{2}}.

$$\frac{1}{2}\left(-2\right)^{2}=\sqrt{8-\left(-2\right)^{2}}$$

Simplify. The value x=-2 satisfies the equation.

$$2=2$$

List all solutions of \frac{x^{2}}{2}=\sqrt{8-x^{2}}.

$$x=2 x=-2$$