Solve -4x^4+5x^2+9=0

$$- 4 x ^ { 4 } + 5 x ^ { 2 } + 9 = 0$$

Solve for x (complex solution)

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-i$$

$$x=i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$-4t^{2}+5t+9=0$$

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, 5 for b, and 9 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-4\right)\times 9}}{-4\times 2}$$

Do the calculations.

$$t=\frac{-5±13}{-8}$$

Solve the equation t=\frac{-5±13}{-8} when ± is plus and when ± is minus.

$$t=-1 t=\frac{9}{4}$$

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

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$-4t^{2}+5t+9=0$$

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, 5 for b, and 9 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-4\right)\times 9}}{-4\times 2}$$

Do the calculations.

$$t=\frac{-5±13}{-8}$$

Solve the equation t=\frac{-5±13}{-8} when ± is plus and when ± is minus.

$$t=-1 t=\frac{9}{4}$$

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

$$x=\frac{3}{2} x=-\frac{3}{2}$$