Solve z^4-17z^2+16=0

$$z ^ { 4 } - 17 z ^ { 2 } + 16 = 0$$

Solve for z

$$z=-4$$

$$z=4$$

$$z=1$$

$$z=-1$$

Steps Using the Quadratic Formula

Substitute t for z^{2}.

$$t^{2}-17t+16=0$$

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, -17 for b, and 16 for c in the quadratic formula.

$$t=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$t=\frac{17±15}{2}$$

Solve the equation t=\frac{17±15}{2} when ± is plus and when ± is minus.

$$t=16 t=1$$

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

$$z=4 z=-4 z=1 z=-1$$