Solve x^4-20x^2+24=0

$$x ^ { 4 } - 20 x ^ { 2 } + 24 = 0$$

Solve for x

$$x=-\sqrt{10-2\sqrt{19}}\approx -1.132343637$$

$$x=\sqrt{10-2\sqrt{19}}\approx 1.132343637$$

$$x=\sqrt{2\sqrt{19}+10}\approx 4.326407041$$

$$x=-\sqrt{2\sqrt{19}+10}\approx -4.326407041$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-20t+24=0$$

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, -20 for b, and 24 for c in the quadratic formula.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 1\times 24}}{2}$$

Do the calculations.

$$t=\frac{20±4\sqrt{19}}{2}$$

Solve the equation t=\frac{20±4\sqrt{19}}{2} when ± is plus and when ± is minus.

$$t=2\sqrt{19}+10 t=10-2\sqrt{19}$$

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

$$x=\frac{\sqrt{8\sqrt{19}+40}}{2} x=-\frac{\sqrt{8\sqrt{19}+40}}{2} x=\frac{\sqrt{40-8\sqrt{19}}}{2} x=-\frac{\sqrt{40-8\sqrt{19}}}{2}$$