Solve x^4+5x^2=14

$$x ^ { 4 } + 5 x ^ { 2 } = 14$$

Solve for x (complex solution)

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

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

$$x=-\sqrt{7}i\approx -0-2.645751311i$$

$$x=\sqrt{7}i\approx 2.645751311i$$

Steps Using the Quadratic Formula

Subtract 14 from both sides.

$$x^{4}+5x^{2}-14=0$$

Substitute t for x^{2}.

$$t^{2}+5t-14=0$$

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 -14 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\times 1\left(-14\right)}}{2}$$

Do the calculations.

$$t=\frac{-5±9}{2}$$

Solve the equation t=\frac{-5±9}{2} when ± is plus and when ± is minus.

$$t=2 t=-7$$

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

$$x=-\sqrt{2} x=\sqrt{2} x=-\sqrt{7}i x=\sqrt{7}i$$

Solve for x

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

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

Steps Using the Quadratic Formula

Subtract 14 from both sides.

$$x^{4}+5x^{2}-14=0$$

Substitute t for x^{2}.

$$t^{2}+5t-14=0$$

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 -14 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\times 1\left(-14\right)}}{2}$$

Do the calculations.

$$t=\frac{-5±9}{2}$$

Solve the equation t=\frac{-5±9}{2} when ± is plus and when ± is minus.

$$t=2 t=-7$$

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

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