Solve 9a^4-97a^2+144=0

$$9 a ^ { 4 } - 97 a ^ { 2 } + 144 = 0$$

Solve for a

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

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

$$a=3$$

$$a=-3$$

Steps Using the Quadratic Formula

Substitute t for a^{2}.

$$9t^{2}-97t+144=0$$

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, -97 for b, and 144 for c in the quadratic formula.

$$t=\frac{-\left(-97\right)±\sqrt{\left(-97\right)^{2}-4\times 9\times 144}}{2\times 9}$$

Do the calculations.

$$t=\frac{97±65}{18}$$

Solve the equation t=\frac{97±65}{18} when ± is plus and when ± is minus.

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

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

$$a=3 a=-3 a=\frac{4}{3} a=-\frac{4}{3}$$