Solve x^4-7x^2-14=4

$$x ^ { 4 } - 7 x ^ { 2 } - 14 = 4$$

Solve for x (complex solution)

$$x=3$$

$$x=-3$$

$$x=-\sqrt{2}i\approx -0-1.414213562i$$

$$x=\sqrt{2}i\approx 1.414213562i$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$x^{4}-7x^{2}-14-4=0$$

Subtract 4 from -14 to get -18.

$$x^{4}-7x^{2}-18=0$$

Substitute t for x^{2}.

$$t^{2}-7t-18=0$$

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, -7 for b, and -18 for c in the quadratic formula.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\left(-18\right)}}{2}$$

Do the calculations.

$$t=\frac{7±11}{2}$$

Solve the equation t=\frac{7±11}{2} when ± is plus and when ± is minus.

$$t=9 t=-2$$

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

$$x=-3 x=3 x=-\sqrt{2}i x=\sqrt{2}i$$

Solve for x

$$x=-3$$

$$x=3$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$x^{4}-7x^{2}-14-4=0$$

Subtract 4 from -14 to get -18.

$$x^{4}-7x^{2}-18=0$$

Substitute t for x^{2}.

$$t^{2}-7t-18=0$$

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, -7 for b, and -18 for c in the quadratic formula.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\left(-18\right)}}{2}$$

Do the calculations.

$$t=\frac{7±11}{2}$$

Solve the equation t=\frac{7±11}{2} when ± is plus and when ± is minus.

$$t=9 t=-2$$

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

$$x=3 x=-3$$