Solve z^4+3z^2-1=0

$$z ^ { 4 } + 3 z ^ { 2 } - 1 = 0$$

Solve for z

$$z=-\frac{\sqrt{2\left(\sqrt{13}-3\right)}}{2}\approx -0.550250523$$

$$z=\frac{\sqrt{2\left(\sqrt{13}-3\right)}}{2}\approx 0.550250523$$

Steps Using the Quadratic Formula

Substitute t for z^{2}.

$$t^{2}+3t-1=0$$

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 1 for a, 3 for b, and -1 for c in the quadratic formula.

$$t=\frac{-3±\sqrt{3^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$t=\frac{-3±\sqrt{13}}{2}$$

Solve the equation t=\frac{-3±\sqrt{13}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{13}-3}{2} t=\frac{-\sqrt{13}-3}{2}$$

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

$$z=\frac{\sqrt{2\sqrt{13}-6}}{2} z=-\frac{\sqrt{2\sqrt{13}-6}}{2}$$