Solve 16x^4+32x^2-153=0

$$16 x ^ { 4 } + 32 x ^ { 2 } - 153 = 0$$

Solve for x (complex solution)

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

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

$$x=-\frac{\sqrt{17}i}{2}\approx -0-2.061552813i$$

$$x=\frac{\sqrt{17}i}{2}\approx 2.061552813i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$16t^{2}+32t-153=0$$

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 16 for a, 32 for b, and -153 for c in the quadratic formula.

$$t=\frac{-32±\sqrt{32^{2}-4\times 16\left(-153\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-32±104}{32}$$

Solve the equation t=\frac{-32±104}{32} when ± is plus and when ± is minus.

$$t=\frac{9}{4} t=-\frac{17}{4}$$

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

$$x=-\frac{3}{2} x=\frac{3}{2} x=-\frac{\sqrt{17}i}{2} x=\frac{\sqrt{17}i}{2}$$

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}.

$$16t^{2}+32t-153=0$$

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 16 for a, 32 for b, and -153 for c in the quadratic formula.

$$t=\frac{-32±\sqrt{32^{2}-4\times 16\left(-153\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-32±104}{32}$$

Solve the equation t=\frac{-32±104}{32} when ± is plus and when ± is minus.

$$t=\frac{9}{4} t=-\frac{17}{4}$$

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

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