Solve a^2+9/a^2=9

$$a ^ { 2 } + \frac { 9 } { a ^ { 2 } } = 9$$

Solve for a

$$a=\frac{-\sqrt{3}-\sqrt{15}}{2}\approx -2.802517077$$

$$a = \frac{\sqrt{3} + \sqrt{15}}{2} \approx 2.802517077$$

$$a = \frac{\sqrt{15} - \sqrt{3}}{2} \approx 1.070466269$$

$$a=\frac{\sqrt{3}-\sqrt{15}}{2}\approx -1.070466269$$

Steps Using the Quadratic Formula

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a^{2}.

$$a^{2}a^{2}+9=9a^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$a^{4}+9=9a^{2}$$

Subtract 9a^{2} from both sides.

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

Substitute t for a^{2}.

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

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

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

Do the calculations.

$$t=\frac{9±3\sqrt{5}}{2}$$

Solve the equation t=\frac{9±3\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{5}+9}{2} t=\frac{9-3\sqrt{5}}{2}$$

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

$$a=\frac{\sqrt{3}+\sqrt{15}}{2} a=-\frac{\sqrt{3}+\sqrt{15}}{2} a=-\frac{\sqrt{3}-\sqrt{15}}{2} a=\frac{\sqrt{3}-\sqrt{15}}{2}$$