Solve x^2+x^2=9/6

$${ x }^{ 2 } + { x }^{ 2 } = \frac{ 9 }{ 6 }$$

Solve for x

$$x=\frac{\sqrt{3}}{2}\approx 0.866025404$$

$$x=-\frac{\sqrt{3}}{2}\approx -0.866025404$$

Steps by Finding Square Root

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

$$2x^{2}=\frac{9}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$2x^{2}=\frac{3}{2}$$

Divide both sides by 2.

$$x^{2}=\frac{\frac{3}{2}}{2}$$

Express \frac{\frac{3}{2}}{2} as a single fraction.

$$x^{2}=\frac{3}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$x^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

$$2x^{2}=\frac{9}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$2x^{2}=\frac{3}{2}$$

Subtract \frac{3}{2} from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and -\frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(-\frac{3}{2}\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(-\frac{3}{2}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(-\frac{3}{2}\right)}}{2\times 2}$$

Multiply -8 times -\frac{3}{2}.

$$x=\frac{0±\sqrt{12}}{2\times 2}$$

Take the square root of 12.

$$x=\frac{0±2\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±2\sqrt{3}}{4}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{4} when ± is plus.

$$x=\frac{\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{4} when ± is minus.

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

The equation is now solved.

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