Solve a^4+1/a^4=119

$$a ^ { 4 } + \frac { 1 } { a ^ { 4 } } = 119$$

Solve for a (complex solution)

$$a\in \frac{-\sqrt{13}-3}{2},-\frac{\sqrt{13}i}{2}-\frac{3}{2}i,\frac{\sqrt{13}i}{2}+\frac{3}{2}i,\frac{\sqrt{13}+3}{2},-\frac{\sqrt{13}i}{2}+\frac{3}{2}i,\frac{3-\sqrt{13}}{2},\frac{\sqrt{13}i}{2}-\frac{3}{2}i,\frac{\sqrt{13}-3}{2}$$

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^{4}.

$$a^{4}a^{4}+1=119a^{4}$$

To multiply powers of the same base, add their exponents. Add 4 and 4 to get 8.

$$a^{8}+1=119a^{4}$$

Subtract 119a^{4} from both sides.

$$a^{8}+1-119a^{4}=0$$

Substitute t for a^{4}.

$$t^{2}-119t+1=0$$

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, -119 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-119\right)±\sqrt{\left(-119\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{119±33\sqrt{13}}{2}$$

Solve the equation t=\frac{119±33\sqrt{13}}{2} when ± is plus and when ± is minus.

$$t=\frac{33\sqrt{13}+119}{2} t=\frac{119-33\sqrt{13}}{2}$$

Since a=t^{4}, the solutions are obtained by solving the equation for each t.

$$a=-\left(\frac{\sqrt{13}i}{2}+\frac{3}{2}i\right) a=-\frac{\sqrt{13}+3}{2} a=\frac{\sqrt{13}i}{2}+\frac{3}{2}i a=\frac{\sqrt{13}+3}{2} a=-\frac{\sqrt{13}i}{2}+\frac{3}{2}i a=\frac{3-\sqrt{13}}{2} a=-\left(-\frac{\sqrt{13}i}{2}+\frac{3}{2}i\right) a=-\frac{3-\sqrt{13}}{2}$$

Solve for a

$$a=\frac{-\sqrt{13}-3}{2}\approx -3.302775638$$

$$a = \frac{\sqrt{13} + 3}{2} \approx 3.302775638$$

$$a=\frac{\sqrt{13}-3}{2}\approx 0.302775638$$

$$a=\frac{3-\sqrt{13}}{2}\approx -0.302775638$$

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^{4}.

$$a^{4}a^{4}+1=119a^{4}$$

To multiply powers of the same base, add their exponents. Add 4 and 4 to get 8.

$$a^{8}+1=119a^{4}$$

Subtract 119a^{4} from both sides.

$$a^{8}+1-119a^{4}=0$$

Substitute t for a^{4}.

$$t^{2}-119t+1=0$$

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, -119 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-119\right)±\sqrt{\left(-119\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{119±33\sqrt{13}}{2}$$

Solve the equation t=\frac{119±33\sqrt{13}}{2} when ± is plus and when ± is minus.

$$t=\frac{33\sqrt{13}+119}{2} t=\frac{119-33\sqrt{13}}{2}$$

Since a=t^{4}, the solutions are obtained by evaluating a=±\sqrt[4]{t} for positive t.

$$a=\frac{\sqrt{13}+3}{2} a=-\frac{\sqrt{13}+3}{2} a=-\frac{3-\sqrt{13}}{2} a=\frac{3-\sqrt{13}}{2}$$