Solve 16x^4-104x^2+73=0

$$16 { x }^{ 4 } -104 { x }^{ 2 } +73=0$$

Solve for x

$$x=-\frac{\sqrt{13-4\sqrt{6}}}{2}\approx -0.894712388$$

$$x=\frac{\sqrt{13-4\sqrt{6}}}{2}\approx 0.894712388$$

$$x = \frac{\sqrt{4 \sqrt{6} + 13}}{2} \approx 2.387360413$$

$$x = -\frac{\sqrt{4 \sqrt{6} + 13}}{2} \approx -2.387360413$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$16t^{2}-104t+73=0$$

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 16 for a, -104 for b, and 73 for c in the quadratic formula.

$$t=\frac{-\left(-104\right)±\sqrt{\left(-104\right)^{2}-4\times 16\times 73}}{2\times 16}$$

Do the calculations.

$$t=\frac{104±32\sqrt{6}}{32}$$

Solve the equation t=\frac{104±32\sqrt{6}}{32} when ± is plus and when ± is minus.

$$t=\sqrt{6}+\frac{13}{4} t=\frac{13}{4}-\sqrt{6}$$

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

$$x=\frac{\sqrt{4\sqrt{6}+13}}{2} x=-\frac{\sqrt{4\sqrt{6}+13}}{2} x=\frac{\sqrt{13-4\sqrt{6}}}{2} x=-\frac{\sqrt{13-4\sqrt{6}}}{2}$$