Solve x^4-5x^2-36=0

$$x ^ { 4 } - 5 x ^ { 2 } - 36 = 0$$

Solve for x (complex solution)

$$x=3$$

$$x=-3$$

$$x=-2i$$

$$x=2i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-5t-36=0$$

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 1 for a, -5 for b, and -36 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-36\right)}}{2}$$

Do the calculations.

$$t=\frac{5±13}{2}$$

Solve the equation t=\frac{5±13}{2} when ± is plus and when ± is minus.

$$t=9 t=-4$$

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

$$x=-3 x=3 x=-2i x=2i$$

Solve for x

$$x=-3$$

$$x=3$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-5t-36=0$$

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 1 for a, -5 for b, and -36 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-36\right)}}{2}$$

Do the calculations.

$$t=\frac{5±13}{2}$$

Solve the equation t=\frac{5±13}{2} when ± is plus and when ± is minus.

$$t=9 t=-4$$

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

$$x=3 x=-3$$