Solve (4+x^2)^2=5

$$( 4 + x ^ { 2 } ) ^ { 2 } = 5$$

Solve for x (complex solution)

$$x=i\sqrt{4-\sqrt{5}}\approx 1.328131026i$$

$$x=-i\sqrt{4-\sqrt{5}}\approx -0-1.328131026i$$

$$x=-i\sqrt{\sqrt{5}+4}\approx -0-2.497212041i$$

$$x=i\sqrt{\sqrt{5}+4}\approx 2.497212041i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+x^{2}\right)^{2}.

$$16+8x^{2}+\left(x^{2}\right)^{2}=5$$

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

$$16+8x^{2}+x^{4}=5$$

Subtract 5 from both sides.

$$16+8x^{2}+x^{4}-5=0$$

Subtract 5 from 16 to get 11.

$$11+8x^{2}+x^{4}=0$$

Substitute t for x^{2}.

$$t^{2}+8t+11=0$$

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, 8 for b, and 11 for c in the quadratic formula.

$$t=\frac{-8±\sqrt{8^{2}-4\times 1\times 11}}{2}$$

Do the calculations.

$$t=\frac{-8±2\sqrt{5}}{2}$$

Solve the equation t=\frac{-8±2\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\sqrt{5}-4 t=-\sqrt{5}-4$$

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

$$x=-i\sqrt{4-\sqrt{5}} x=i\sqrt{4-\sqrt{5}} x=-i\sqrt{\sqrt{5}+4} x=i\sqrt{\sqrt{5}+4}$$