Solve 1+x^4=5x^2

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

Solve for x

$$x=\frac{\sqrt{3}-\sqrt{7}}{2}\approx -0.456850252$$

$$x=\frac{\sqrt{7}-\sqrt{3}}{2}\approx 0.456850252$$

$$x = \frac{\sqrt{3} + \sqrt{7}}{2} \approx 2.188901059$$

$$x=\frac{-\sqrt{3}-\sqrt{7}}{2}\approx -2.188901059$$

Steps Using the Quadratic Formula

Subtract 5x^{2} from both sides.

$$1+x^{4}-5x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-5t+1=0$$

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, -5 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{5±\sqrt{21}}{2}$$

Solve the equation t=\frac{5±\sqrt{21}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{21}+5}{2} t=\frac{5-\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}-\sqrt{7}}{2} x=\frac{\sqrt{3}-\sqrt{7}}{2}$$