Solve 4x^4+32x^2-36=0

$$4 { x }^{ 4 } +32 { x }^{ 2 } -36=0$$

Solve for x (complex solution)

$$x=1$$

$$x=-1$$

$$x=-3i$$

$$x=3i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+32t-36=0$$

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 4 for a, 32 for b, and -36 for c in the quadratic formula.

$$t=\frac{-32±\sqrt{32^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{-32±40}{8}$$

Solve the equation t=\frac{-32±40}{8} when ± is plus and when ± is minus.

$$t=1 t=-9$$

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

$$x=-1 x=1 x=-3i x=3i$$

Solve for x

$$x=-1$$

$$x=1$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+32t-36=0$$

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 4 for a, 32 for b, and -36 for c in the quadratic formula.

$$t=\frac{-32±\sqrt{32^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{-32±40}{8}$$

Solve the equation t=\frac{-32±40}{8} when ± is plus and when ± is minus.

$$t=1 t=-9$$

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

$$x=1 x=-1$$