Solve (z^2-2)^2-36=0

$$( z ^ { 2 } - 2 ) ^ { 2 } - 36 = 0$$

Solve for z

$$z=-2\sqrt{2}\approx -2.828427125$$

$$z=2\sqrt{2}\approx 2.828427125$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(z^{2}-2\right)^{2}.

$$\left(z^{2}\right)^{2}-4z^{2}+4-36=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$z^{4}-4z^{2}+4-36=0$$

Subtract 36 from 4 to get -32.

$$z^{4}-4z^{2}-32=0$$

Substitute t for z^{2}.

$$t^{2}-4t-32=0$$

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, -4 for b, and -32 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-32\right)}}{2}$$

Do the calculations.

$$t=\frac{4±12}{2}$$

Solve the equation t=\frac{4±12}{2} when ± is plus and when ± is minus.

$$t=8 t=-4$$

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

$$z=2\sqrt{2} z=-2\sqrt{2}$$