Solve 3x^4+17x^2-6=0

$$3 x ^ { 4 } + 17 x ^ { 2 } - 6 = 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3}}{3}\approx 0.577350269$$

$$x=-\frac{\sqrt{3}}{3}\approx -0.577350269$$

$$x=-\sqrt{6}i\approx -0-2.449489743i$$

$$x=\sqrt{6}i\approx 2.449489743i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$3t^{2}+17t-6=0$$

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 3 for a, 17 for b, and -6 for c in the quadratic formula.

$$t=\frac{-17±\sqrt{17^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Do the calculations.

$$t=\frac{-17±19}{6}$$

Solve the equation t=\frac{-17±19}{6} when ± is plus and when ± is minus.

$$t=\frac{1}{3} t=-6$$

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

$$x=-\frac{\sqrt{3}}{3} x=\frac{\sqrt{3}}{3} x=-\sqrt{6}i x=\sqrt{6}i$$

Solve for x

$$x=-\frac{\sqrt{3}}{3}\approx -0.577350269$$

$$x=\frac{\sqrt{3}}{3}\approx 0.577350269$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$3t^{2}+17t-6=0$$

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 3 for a, 17 for b, and -6 for c in the quadratic formula.

$$t=\frac{-17±\sqrt{17^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

Do the calculations.

$$t=\frac{-17±19}{6}$$

Solve the equation t=\frac{-17±19}{6} when ± is plus and when ± is minus.

$$t=\frac{1}{3} t=-6$$

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

$$x=\frac{\sqrt{3}}{3} x=-\frac{\sqrt{3}}{3}$$