Solve x^-2+6x^-1+8=0

$$x ^ { - 2 } + 6 x ^ { - 1 } + 8 = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-\frac{1}{4}=-0.25$$

Steps Using Factoring By Grouping

Reorder the terms.

$$8+6\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 8+6\times 1+xx^{-2}=0$$

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

$$x\times 8+6\times 1+x^{-1}=0$$

Multiply 6 and 1 to get 6.

$$x\times 8+6+x^{-1}=0$$

Reorder the terms.

$$8x+6+\frac{1}{x}=0$$

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

$$8xx+x\times 6+1=0$$

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

$$8x^{2}+x\times 6+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=6 ab=8\times 1=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite 8x^{2}+6x+1 as \left(8x^{2}+2x\right)+\left(4x+1\right).

$$\left(8x^{2}+2x\right)+\left(4x+1\right)$$

Factor out 2x in 8x^{2}+2x.

$$2x\left(4x+1\right)+4x+1$$

Factor out common term 4x+1 by using distributive property.

$$\left(4x+1\right)\left(2x+1\right)$$

To find equation solutions, solve 4x+1=0 and 2x+1=0.

$$x=-\frac{1}{4} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Reorder the terms.

$$8+6\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 8+6\times 1+xx^{-2}=0$$

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

$$x\times 8+6\times 1+x^{-1}=0$$

Multiply 6 and 1 to get 6.

$$x\times 8+6+x^{-1}=0$$

Reorder the terms.

$$8x+6+\frac{1}{x}=0$$

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

$$8xx+x\times 6+1=0$$

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

$$8x^{2}+x\times 6+1=0$$

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

$$8x^{2}+6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 8}}{2\times 8}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-6±\sqrt{36-32}}{2\times 8}$$

Add 36 to -32.

$$x=\frac{-6±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

$$x=\frac{-6±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-6±2}{16}$$

Now solve the equation x=\frac{-6±2}{16} when ± is plus. Add -6 to 2.

$$x=-\frac{4}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{4}$$

Now solve the equation x=\frac{-6±2}{16} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{16}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{1}{4} x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$x^{-2}+6x^{-1}=-8$$

Reorder the terms.

$$6\times \frac{1}{x}+x^{-2}=-8$$

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

$$6\times 1+xx^{-2}=-8x$$

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

$$6\times 1+x^{-1}=-8x$$

Multiply 6 and 1 to get 6.

$$6+x^{-1}=-8x$$

Add 8x to both sides.

$$6+x^{-1}+8x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$x^{-1}+8x=-6$$

Reorder the terms.

$$8x+\frac{1}{x}=-6$$

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

$$8xx+1=-6x$$

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

$$8x^{2}+1=-6x$$

Add 6x to both sides.

$$8x^{2}+1+6x=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}+6x=-1$$

Divide both sides by 8.

$$\frac{8x^{2}+6x}{8}=-\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{6}{8}x=-\frac{1}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{4}x=-\frac{1}{8}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=-\frac{1}{8}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=-\frac{1}{8}+\frac{9}{64}$$

Add -\frac{1}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{1}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{1}{8} x+\frac{3}{8}=-\frac{1}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=-\frac{1}{4} x=-\frac{1}{2}$$