Solve 4x^4-65x^2+16=0

$$4 x ^ { 4 } - 65 x ^ { 2 } + 16 = 0$$

Solve for x

$$x=-4$$

$$x=4$$

$$x=\frac{1}{2}=0.5$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}-65t+16=0$$

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 4 for a, -65 for b, and 16 for c in the quadratic formula.

$$t=\frac{-\left(-65\right)±\sqrt{\left(-65\right)^{2}-4\times 4\times 16}}{2\times 4}$$

Do the calculations.

$$t=\frac{65±63}{8}$$

Solve the equation t=\frac{65±63}{8} when ± is plus and when ± is minus.

$$t=16 t=\frac{1}{4}$$

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

$$x=4 x=-4 x=\frac{1}{2} x=-\frac{1}{2}$$