Solve -2/5(x^2+1/2)^2=-3/8

$$- \frac { 2 } { 5 } ( x ^ { 2 } + \frac { 1 } { 2 } ) ^ { 2 } = - \frac { 3 } { 8 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0.684284909$$

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0.684284909$$

$$x=-\frac{i\sqrt{\sqrt{15}+2}}{2}\approx -0-1.211711945i$$

$$x=\frac{i\sqrt{\sqrt{15}+2}}{2}\approx 1.211711945i$$

Steps Using the Quadratic Formula

Multiply both sides by -\frac{5}{2}, the reciprocal of -\frac{2}{5}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

Multiply -\frac{3}{8} and -\frac{5}{2} to get \frac{15}{16}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

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

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Subtract \frac{15}{16} from both sides.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

Subtract \frac{15}{16} from \frac{1}{4} to get -\frac{11}{16}.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

Substitute t for x^{2}.

$$t^{2}+t-\frac{11}{16}=0$$

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 1 for a, 1 for b, and -\frac{11}{16} for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Do the calculations.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

Solve the equation t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2} x=\frac{\sqrt{\sqrt{15}-2}}{2} x=-\frac{i\sqrt{\sqrt{15}+2}}{2} x=\frac{i\sqrt{\sqrt{15}+2}}{2}$$

Solve for x

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0.684284909$$

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0.684284909$$

Steps Using the Quadratic Formula

Multiply both sides by -\frac{5}{2}, the reciprocal of -\frac{2}{5}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

Multiply -\frac{3}{8} and -\frac{5}{2} to get \frac{15}{16}.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

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

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Subtract \frac{15}{16} from both sides.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

Subtract \frac{15}{16} from \frac{1}{4} to get -\frac{11}{16}.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

Substitute t for x^{2}.

$$t^{2}+t-\frac{11}{16}=0$$

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 1 for a, 1 for b, and -\frac{11}{16} for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Do the calculations.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

Solve the equation t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

$$x=\frac{\sqrt{\sqrt{15}-2}}{2} x=-\frac{\sqrt{\sqrt{15}-2}}{2}$$