Solve -3x^-2+5=8x^-1

$$- 3 x ^ { - 2 } + 5 = 8 x ^ { - 1 }$$

Solve for x

$$x=\frac{4-\sqrt{31}}{5}\approx -0.313552873$$

$$x = \frac{\sqrt{31} + 4}{5} \approx 1.913552873$$

Steps Using the Quadratic Formula

Subtract 8x^{-1} from both sides.

$$-3x^{-2}+5-8x^{-1}=0$$

Reorder the terms.

$$5-8\times \frac{1}{x}-3x^{-2}=0$$

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

$$x\times 5-8-3x^{-2}x=0$$

To multiply powers of the same base, add their exponents. Add -2 and 1 to get -1.

$$x\times 5-8-3x^{-1}=0$$

Reorder the terms.

$$5x-8-3\times \frac{1}{x}=0$$

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

$$5xx+x\left(-8\right)-3=0$$

Multiply x and x to get x^{2}.

$$5x^{2}+x\left(-8\right)-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$5x^{2}-8x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64+60}}{2\times 5}$$

Add 64 to 60.

$$x=\frac{-\left(-8\right)±\sqrt{124}}{2\times 5}$$

Take the square root of 124.

$$x=\frac{-\left(-8\right)±2\sqrt{31}}{2\times 5}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{31}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{8±2\sqrt{31}}{10}$$

Now solve the equation x=\frac{8±2\sqrt{31}}{10} when ± is plus. Add 8 to 2\sqrt{31}.

$$x=\frac{2\sqrt{31}+8}{10}$$

Divide 8+2\sqrt{31} by 10.

$$x=\frac{\sqrt{31}+4}{5}$$

Now solve the equation x=\frac{8±2\sqrt{31}}{10} when ± is minus. Subtract 2\sqrt{31} from 8.

$$x=\frac{8-2\sqrt{31}}{10}$$

Divide 8-2\sqrt{31} by 10.

$$x=\frac{4-\sqrt{31}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{31}+4}{5} x=\frac{4-\sqrt{31}}{5}$$

Steps for Completing the Square

Subtract 8x^{-1} from both sides.

$$-3x^{-2}+5-8x^{-1}=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{-2}-8x^{-1}=-5$$

Reorder the terms.

$$-8\times \frac{1}{x}-3x^{-2}=-5$$

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

$$-8-3x^{-2}x=-5x$$

To multiply powers of the same base, add their exponents. Add -2 and 1 to get -1.

$$-8-3x^{-1}=-5x$$

Add 5x to both sides.

$$-8-3x^{-1}+5x=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$-3x^{-1}+5x=8$$

Reorder the terms.

$$5x-3\times \frac{1}{x}=8$$

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

$$5xx-3=8x$$

Multiply x and x to get x^{2}.

$$5x^{2}-3=8x$$

Subtract 8x from both sides.

$$5x^{2}-3-8x=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$5x^{2}-8x=3$$

Divide both sides by 5.

$$\frac{5x^{2}-8x}{5}=\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{8}{5}x=\frac{3}{5}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=\frac{3}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{3}{5}+\frac{16}{25}$$

Add \frac{3}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{31}{25}$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{31}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{31}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{\sqrt{31}}{5} x-\frac{4}{5}=-\frac{\sqrt{31}}{5}$$

Add \frac{4}{5} to both sides of the equation.

$$x=\frac{\sqrt{31}+4}{5} x=\frac{4-\sqrt{31}}{5}$$