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

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

Solve for x (complex solution)

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

$$x=0$$

$$x=\frac{-\sqrt{7}i-1}{4}\approx -0.25-0.661437828i$$

$$x=\frac{-1+\sqrt{7}i}{4}\approx -0.25+0.661437828i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+t-1=0$$

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, 1 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-1±\sqrt{17}}{8}$$

Solve the equation t=\frac{-1±\sqrt{17}}{8} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{17}-1}{8} t=\frac{-\sqrt{17}-1}{8}$$

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

$$x=-\sqrt{\frac{\sqrt{17}-1}{8}} x=\sqrt{\frac{\sqrt{17}-1}{8}} x=-i\sqrt{\frac{\sqrt{17}+1}{8}} x=i\sqrt{\frac{\sqrt{17}+1}{8}}$$

Solve for x

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

$$x=0$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+t-1=0$$

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, 1 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-1±\sqrt{17}}{8}$$

Solve the equation t=\frac{-1±\sqrt{17}}{8} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{17}-1}{8} t=\frac{-\sqrt{17}-1}{8}$$

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

$$x=\frac{\sqrt{\frac{\sqrt{17}-1}{2}}}{2} x=-\frac{\sqrt{\frac{\sqrt{17}-1}{2}}}{2}$$