Solve a^4-25/4a^2+9=0

$$a ^ { 4 } - \frac { 25 } { 4 } a ^ { 2 } + 9 = 0$$

Solve for a

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

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

$$a=2$$

$$a=-2$$

Steps Using the Quadratic Formula

Substitute t for a^{2}.

$$t^{2}-\frac{25}{4}t+9=0$$

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 1 for a, -\frac{25}{4} for b, and 9 for c in the quadratic formula.

$$t=\frac{-\left(-\frac{25}{4}\right)±\sqrt{\left(-\frac{25}{4}\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$t=\frac{\frac{25}{4}±\frac{7}{4}}{2}$$

Solve the equation t=\frac{\frac{25}{4}±\frac{7}{4}}{2} when ± is plus and when ± is minus.

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

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

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