Solve 2k^4+14k^2-1=0

$$2 k ^ { 4 } + 14 k ^ { 2 } - 1 = 0$$

Solve for k (complex solution)

$$k=\frac{i\sqrt{2\left(\sqrt{51}+7\right)}}{2}\approx 2.659081461i$$

$$k=-\frac{i\sqrt{2\left(\sqrt{51}+7\right)}}{2}\approx -0-2.659081461i$$

$$k=-\frac{\sqrt{2\left(\sqrt{51}-7\right)}}{2}\approx -0.265921444$$

$$k=\frac{\sqrt{2\left(\sqrt{51}-7\right)}}{2}\approx 0.265921444$$

Steps Using the Quadratic Formula

Substitute t for k^{2}.

$$2t^{2}+14t-1=0$$

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 2 for a, 14 for b, and -1 for c in the quadratic formula.

$$t=\frac{-14±\sqrt{14^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Do the calculations.

$$t=\frac{-14±2\sqrt{51}}{4}$$

Solve the equation t=\frac{-14±2\sqrt{51}}{4} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{51}-7}{2} t=\frac{-\sqrt{51}-7}{2}$$

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

$$k=-\sqrt{\frac{\sqrt{51}-7}{2}} k=\sqrt{\frac{\sqrt{51}-7}{2}} k=-i\sqrt{\frac{\sqrt{51}+7}{2}} k=i\sqrt{\frac{\sqrt{51}+7}{2}}$$

Solve for k

$$k=-\frac{\sqrt{2\left(\sqrt{51}-7\right)}}{2}\approx -0.265921444$$

$$k=\frac{\sqrt{2\left(\sqrt{51}-7\right)}}{2}\approx 0.265921444$$

Steps Using the Quadratic Formula

Substitute t for k^{2}.

$$2t^{2}+14t-1=0$$

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 2 for a, 14 for b, and -1 for c in the quadratic formula.

$$t=\frac{-14±\sqrt{14^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Do the calculations.

$$t=\frac{-14±2\sqrt{51}}{4}$$

Solve the equation t=\frac{-14±2\sqrt{51}}{4} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{51}-7}{2} t=\frac{-\sqrt{51}-7}{2}$$

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

$$k=\frac{\sqrt{2\sqrt{51}-14}}{2} k=-\frac{\sqrt{2\sqrt{51}-14}}{2}$$