Solve x^2+4^2=(24x^2)^2

$$x ^ { 2 } + 4 ^ { 2 } = ( 24 x ^ { 2 } ) ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{36865}-1\right)}}{48}\approx 0.407186532i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{36865}-1\right)}}{48}\approx -0-0.407186532i$$

$$x=-\frac{\sqrt{2\left(\sqrt{36865}+1\right)}}{48}\approx -0.409312818$$

$$x=\frac{\sqrt{2\left(\sqrt{36865}+1\right)}}{48}\approx 0.409312818$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$x^{2}+16=\left(24x^{2}\right)^{2}$$

Expand \left(24x^{2}\right)^{2}.

$$x^{2}+16=24^{2}\left(x^{2}\right)^{2}$$

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

$$x^{2}+16=24^{2}x^{4}$$

Calculate 24 to the power of 2 and get 576.

$$x^{2}+16=576x^{4}$$

Subtract 576x^{4} from both sides.

$$x^{2}+16-576x^{4}=0$$

Substitute t for x^{2}.

$$-576t^{2}+t+16=0$$

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 -576 for a, 1 for b, and 16 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-576\right)\times 16}}{-576\times 2}$$

Do the calculations.

$$t=\frac{-1±\sqrt{36865}}{-1152}$$

Solve the equation t=\frac{-1±\sqrt{36865}}{-1152} when ± is plus and when ± is minus.

$$t=\frac{1-\sqrt{36865}}{1152} t=\frac{\sqrt{36865}+1}{1152}$$

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

$$x=-i\sqrt{-\frac{1-\sqrt{36865}}{1152}} x=i\sqrt{-\frac{1-\sqrt{36865}}{1152}} x=-\sqrt{\frac{\sqrt{36865}+1}{1152}} x=\sqrt{\frac{\sqrt{36865}+1}{1152}}$$

Solve for x

$$x=-\frac{\sqrt{2\left(\sqrt{36865}+1\right)}}{48}\approx -0.409312818$$

$$x=\frac{\sqrt{2\left(\sqrt{36865}+1\right)}}{48}\approx 0.409312818$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$x^{2}+16=\left(24x^{2}\right)^{2}$$

Expand \left(24x^{2}\right)^{2}.

$$x^{2}+16=24^{2}\left(x^{2}\right)^{2}$$

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

$$x^{2}+16=24^{2}x^{4}$$

Calculate 24 to the power of 2 and get 576.

$$x^{2}+16=576x^{4}$$

Subtract 576x^{4} from both sides.

$$x^{2}+16-576x^{4}=0$$

Substitute t for x^{2}.

$$-576t^{2}+t+16=0$$

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 -576 for a, 1 for b, and 16 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-576\right)\times 16}}{-576\times 2}$$

Do the calculations.

$$t=\frac{-1±\sqrt{36865}}{-1152}$$

Solve the equation t=\frac{-1±\sqrt{36865}}{-1152} when ± is plus and when ± is minus.

$$t=\frac{1-\sqrt{36865}}{1152} t=\frac{\sqrt{36865}+1}{1152}$$

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

$$x=\frac{\sqrt{\frac{\sqrt{36865}+1}{32}}}{6} x=-\frac{\sqrt{\frac{\sqrt{36865}+1}{32}}}{6}$$