Solve x^4-5x^2+3=0

$${ x }^{ 4 } -5 { x }^{ 2 } +3 = 0$$

Solve for x

$$x=-\frac{\sqrt{10-2\sqrt{13}}}{2}\approx -0.834999618$$

$$x=\frac{\sqrt{10-2\sqrt{13}}}{2}\approx 0.834999618$$

$$x = \frac{\sqrt{2 \sqrt{13} + 10}}{2} \approx 2.074313293$$

$$x = -\frac{\sqrt{2 \sqrt{13} + 10}}{2} \approx -2.074313293$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-5t+3=0$$

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 3 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$t=\frac{5±\sqrt{13}}{2}$$

Solve the equation t=\frac{5±\sqrt{13}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{13}+5}{2} t=\frac{5-\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{13}+10}}{2} x=-\frac{\sqrt{2\sqrt{13}+10}}{2} x=\frac{\sqrt{10-2\sqrt{13}}}{2} x=-\frac{\sqrt{10-2\sqrt{13}}}{2}$$