Solve 16x^4+7x^2-50=0

$$16 x ^ { 4 } + 7 x ^ { 2 } - 50 = 0$$

Solve for x (complex solution)

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

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

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

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$16t^{2}+7t-50=0$$

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, 7 for b, and -50 for c in the quadratic formula.

$$t=\frac{-7±\sqrt{7^{2}-4\times 16\left(-50\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-7±57}{32}$$

Solve the equation t=\frac{-7±57}{32} when ± is plus and when ± is minus.

$$t=\frac{25}{16} t=-2$$

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

$$x=-\frac{5}{4} x=\frac{5}{4} x=-\sqrt{2}i x=\sqrt{2}i$$

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$16t^{2}+7t-50=0$$

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, 7 for b, and -50 for c in the quadratic formula.

$$t=\frac{-7±\sqrt{7^{2}-4\times 16\left(-50\right)}}{2\times 16}$$

Do the calculations.

$$t=\frac{-7±57}{32}$$

Solve the equation t=\frac{-7±57}{32} when ± is plus and when ± is minus.

$$t=\frac{25}{16} t=-2$$

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

$$x=\frac{5}{4} x=-\frac{5}{4}$$