Solve x^4-23x^2+36=0

$$x ^ { 4 } - 23 x ^ { 2 } + 36 = 0$$

Solve for x

$$x=\frac{\sqrt{11}-\sqrt{35}}{2}\approx -1.299727496$$

$$x = \frac{\sqrt{35} - \sqrt{11}}{2} \approx 1.299727496$$

$$x = \frac{\sqrt{11} + \sqrt{35}}{2} \approx 4.616352287$$

$$x=\frac{-\sqrt{11}-\sqrt{35}}{2}\approx -4.616352287$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-23t+36=0$$

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, -23 for b, and 36 for c in the quadratic formula.

$$t=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$t=\frac{23±\sqrt{385}}{2}$$

Solve the equation t=\frac{23±\sqrt{385}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{385}+23}{2} t=\frac{23-\sqrt{385}}{2}$$

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

$$x=\frac{\sqrt{11}+\sqrt{35}}{2} x=-\frac{\sqrt{11}+\sqrt{35}}{2} x=-\frac{\sqrt{11}-\sqrt{35}}{2} x=\frac{\sqrt{11}-\sqrt{35}}{2}$$