Solve 4x^4+19x^2-5=0

$$4 x ^ { 4 } + 19 x ^ { 2 } - 5 = 0$$

Solve for x (complex solution)

$$x=\frac{1}{2}=0.5$$

$$x=-\frac{1}{2}=-0.5$$

$$x=-\sqrt{5}i\approx -0-2.236067977i$$

$$x=\sqrt{5}i\approx 2.236067977i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+19t-5=0$$

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, 19 for b, and -5 for c in the quadratic formula.

$$t=\frac{-19±\sqrt{19^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{-19±21}{8}$$

Solve the equation t=\frac{-19±21}{8} when ± is plus and when ± is minus.

$$t=\frac{1}{4} t=-5$$

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

$$x=-\frac{1}{2} x=\frac{1}{2} x=-\sqrt{5}i x=\sqrt{5}i$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$4t^{2}+19t-5=0$$

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, 19 for b, and -5 for c in the quadratic formula.

$$t=\frac{-19±\sqrt{19^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Do the calculations.

$$t=\frac{-19±21}{8}$$

Solve the equation t=\frac{-19±21}{8} when ± is plus and when ± is minus.

$$t=\frac{1}{4} t=-5$$

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

$$x=\frac{1}{2} x=-\frac{1}{2}$$