Solve x^2+1/x^2=83.text{Findthevalueof}x^3-1/x^3

$$x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = 83 . \text { Find the value of } x ^ { 3 } - \frac { 1 } { x ^ { 3 } }$$

Solve for x

$$x=\frac{-\sqrt{85}-9}{2}\approx -9.109772229$$

$$x = \frac{\sqrt{85} + 9}{2} \approx 9.109772229$$

$$x=\frac{\sqrt{85}-9}{2}\approx 0.109772229$$

$$x=\frac{9-\sqrt{85}}{2}\approx -0.109772229$$

Steps Using the Quadratic Formula

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

$$x^{2}x^{2}+1=83x^{2}$$

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

$$x^{4}+1=83x^{2}$$

Subtract 83x^{2} from both sides.

$$x^{4}+1-83x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-83t+1=0$$

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, -83 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-83\right)±\sqrt{\left(-83\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{83±9\sqrt{85}}{2}$$

Solve the equation t=\frac{83±9\sqrt{85}}{2} when ± is plus and when ± is minus.

$$t=\frac{9\sqrt{85}+83}{2} t=\frac{83-9\sqrt{85}}{2}$$

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

$$x=\frac{\sqrt{85}+9}{2} x=-\frac{\sqrt{85}+9}{2} x=-\frac{9-\sqrt{85}}{2} x=\frac{9-\sqrt{85}}{2}$$