Solve 4x^4-19x^2+18=0

$$4 x ^ { 4 } - 19 x ^ { 2 } + 18 = 0$$

Solve for x

$$x = -\frac{\sqrt{38 - 2 \sqrt{73}}}{4} \approx -1.143240802$$

$$x = \frac{\sqrt{38 - 2 \sqrt{73}}}{4} \approx 1.143240802$$

$$x = \frac{\sqrt{2 \sqrt{73} + 38}}{4} \approx 1.855532395$$

$$x = -\frac{\sqrt{2 \sqrt{73} + 38}}{4} \approx -1.855532395$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}-19t+18=0$$

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, -19 for b, and 18 for c in the quadratic formula.

$$t=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 4\times 18}}{2\times 4}$$

Do the calculations.

$$t=\frac{19±\sqrt{73}}{8}$$

Solve the equation t=\frac{19±\sqrt{73}}{8} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{73}+19}{8} t=\frac{19-\sqrt{73}}{8}$$

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

$$x=\frac{\sqrt{\frac{\sqrt{73}+19}{2}}}{2} x=-\frac{\sqrt{\frac{\sqrt{73}+19}{2}}}{2} x=\frac{\sqrt{\frac{19-\sqrt{73}}{2}}}{2} x=-\frac{\sqrt{\frac{19-\sqrt{73}}{2}}}{2}$$