Solve 9x^4=25x^2-16

$$9 x ^ { 4 } = 25 x ^ { 2 } - 16$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Subtract 25x^{2} from both sides.

$$9x^{4}-25x^{2}=-16$$

Add 16 to both sides.

$$9x^{4}-25x^{2}+16=0$$

Substitute t for x^{2}.

$$9t^{2}-25t+16=0$$

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 9 for a, -25 for b, and 16 for c in the quadratic formula.

$$t=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 9\times 16}}{2\times 9}$$

Do the calculations.

$$t=\frac{25±7}{18}$$

Solve the equation t=\frac{25±7}{18} when ± is plus and when ± is minus.

$$t=\frac{16}{9} t=1$$

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

$$x=\frac{4}{3} x=-\frac{4}{3} x=1 x=-1$$