Solve 4(x^4+1)=5x^2

$$4 ( x ^ { 4 } + 1 ) = 5 x ^ { 2 }$$

Solve for x (complex solution)

$$x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0.901387819+0.433012702i$$

$$x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0.901387819-0.433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0.901387819-0.433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0.901387819+0.433012702i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by x^{4}+1.

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

Subtract 5x^{2} from both sides.

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

Substitute t for x^{2}.

$$4t^{2}-5t+4=0$$

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 4 for a, -5 for b, and 4 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Do the calculations.

$$t=\frac{5±\sqrt{-39}}{8}$$

Solve the equation t=\frac{5±\sqrt{-39}}{8} when ± is plus and when ± is minus.

$$t=\frac{5+\sqrt{39}i}{8} t=\frac{-\sqrt{39}i+5}{8}$$

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

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}} x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}$$