Solve (1/2x^2+2/3)^2=1/4x^4+1/2-x+frac{2}{3}x^2

$$( \frac { 1 } { 2 } x ^ { 2 } + \frac { 2 } { 3 } ) ^ { 2 } = \frac { 1 } { 4 } x ^ { 4 } + \frac { 1 } { 2 } - x + \frac { 2 } { 3 } x ^ { 2 }$$

Solve for x

$$x=\frac{1}{18}\approx 0.055555556$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x^{2}+\frac{2}{3}\right)^{2}.

$$\frac{1}{4}\left(x^{2}\right)^{2}+\frac{2}{3}x^{2}+\frac{4}{9}=\frac{1}{4}x^{4}+\frac{1}{2}-x+\frac{2}{3}x^{2}$$

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

$$\frac{1}{4}x^{4}+\frac{2}{3}x^{2}+\frac{4}{9}=\frac{1}{4}x^{4}+\frac{1}{2}-x+\frac{2}{3}x^{2}$$

Subtract \frac{1}{4}x^{4} from both sides.

$$\frac{1}{4}x^{4}+\frac{2}{3}x^{2}+\frac{4}{9}-\frac{1}{4}x^{4}=\frac{1}{2}-x+\frac{2}{3}x^{2}$$

Combine \frac{1}{4}x^{4} and -\frac{1}{4}x^{4} to get 0.

$$\frac{2}{3}x^{2}+\frac{4}{9}=\frac{1}{2}-x+\frac{2}{3}x^{2}$$

Add x to both sides.

$$\frac{2}{3}x^{2}+\frac{4}{9}+x=\frac{1}{2}+\frac{2}{3}x^{2}$$

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

$$\frac{2}{3}x^{2}+\frac{4}{9}+x-\frac{2}{3}x^{2}=\frac{1}{2}$$

Combine \frac{2}{3}x^{2} and -\frac{2}{3}x^{2} to get 0.

$$\frac{4}{9}+x=\frac{1}{2}$$

Subtract \frac{4}{9} from both sides.

$$x=\frac{1}{2}-\frac{4}{9}$$

Subtract \frac{4}{9} from \frac{1}{2} to get \frac{1}{18}.

$$x=\frac{1}{18}$$