Solve x^-2+9x^-1+8=0

$$x ^ { - 2 } + 9 x ^ { - 1 } + 8 = 0$$

Solve for x

$$x=-1$$

$$x=-\frac{1}{8}=-0.125$$

Steps Using Factoring By Grouping

Reorder the terms.

$$8+9\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 8+9\times 1+xx^{-2}=0$$

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

$$x\times 8+9\times 1+x^{-1}=0$$

Multiply 9 and 1 to get 9.

$$x\times 8+9+x^{-1}=0$$

Reorder the terms.

$$8x+9+\frac{1}{x}=0$$

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

$$8xx+x\times 9+1=0$$

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

$$8x^{2}+x\times 9+1=0$$

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.

$$a+b=9 ab=8\times 1=8$$

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 9.

$$a=1 b=8$$

Rewrite 8x^{2}+9x+1 as \left(8x^{2}+x\right)+\left(8x+1\right).

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

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

$$x\left(8x+1\right)+8x+1$$

Factor out common term 8x+1 by using distributive property.

$$\left(8x+1\right)\left(x+1\right)$$

To find equation solutions, solve 8x+1=0 and x+1=0.

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

Steps Using the Quadratic Formula

Reorder the terms.

$$8+9\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 8+9\times 1+xx^{-2}=0$$

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

$$x\times 8+9\times 1+x^{-1}=0$$

Multiply 9 and 1 to get 9.

$$x\times 8+9+x^{-1}=0$$

Reorder the terms.

$$8x+9+\frac{1}{x}=0$$

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

$$8xx+x\times 9+1=0$$

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

$$8x^{2}+x\times 9+1=0$$

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.

$$8x^{2}+9x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 9 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 8}}{2\times 8}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-9±\sqrt{81-32}}{2\times 8}$$

Add 81 to -32.

$$x=\frac{-9±\sqrt{49}}{2\times 8}$$

Take the square root of 49.

$$x=\frac{-9±7}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-9±7}{16}$$

Now solve the equation x=\frac{-9±7}{16} when ± is plus. Add -9 to 7.

$$x=-\frac{2}{16}$$

Reduce the fraction \frac{-2}{16} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-9±7}{16} when ± is minus. Subtract 7 from -9.

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

Divide -16 by 16.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{-2}+9x^{-1}=-8$$

Reorder the terms.

$$9\times \frac{1}{x}+x^{-2}=-8$$

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

$$9\times 1+xx^{-2}=-8x$$

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

$$9\times 1+x^{-1}=-8x$$

Multiply 9 and 1 to get 9.

$$9+x^{-1}=-8x$$

Add 8x to both sides.

$$9+x^{-1}+8x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$x^{-1}+8x=-9$$

Reorder the terms.

$$8x+\frac{1}{x}=-9$$

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

$$8xx+1=-9x$$

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

$$8x^{2}+1=-9x$$

Add 9x to both sides.

$$8x^{2}+1+9x=0$$

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

$$8x^{2}+9x=-1$$

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide \frac{9}{8}, the coefficient of the x term, by 2 to get \frac{9}{16}. Then add the square of \frac{9}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{8}x+\left(\frac{9}{16}\right)^{2}=-\frac{1}{8}+\left(\frac{9}{16}\right)^{2}$$

Square \frac{9}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{8}x+\frac{81}{256}=-\frac{1}{8}+\frac{81}{256}$$

Add -\frac{1}{8} to \frac{81}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{8}x+\frac{81}{256}=\frac{49}{256}$$

Factor x^{2}+\frac{9}{8}x+\frac{81}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{16}\right)^{2}=\frac{49}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{16}\right)^{2}}=\sqrt{\frac{49}{256}}$$

Simplify.

$$x+\frac{9}{16}=\frac{7}{16} x+\frac{9}{16}=-\frac{7}{16}$$

Subtract \frac{9}{16} from both sides of the equation.

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