Solve x^-2+8x^-1+7=0

$$x ^ { - 2 } + 8 x ^ { - 1 } + 7 = 0$$

Solve for x

$$x=-1$$

$$x=-\frac{1}{7}\approx -0.142857143$$

Steps Using Factoring By Grouping

Reorder the terms.

$$7+8\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 7+8\times 1+xx^{-2}=0$$

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

$$x\times 7+8\times 1+x^{-1}=0$$

Multiply 8 and 1 to get 8.

$$x\times 7+8+x^{-1}=0$$

Reorder the terms.

$$7x+8+\frac{1}{x}=0$$

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

$$7xx+x\times 8+1=0$$

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

$$7x^{2}+x\times 8+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=8 ab=7\times 1=7$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=7$$

Rewrite 7x^{2}+8x+1 as \left(7x^{2}+x\right)+\left(7x+1\right).

$$\left(7x^{2}+x\right)+\left(7x+1\right)$$

Factor out x in 7x^{2}+x.

$$x\left(7x+1\right)+7x+1$$

Factor out common term 7x+1 by using distributive property.

$$\left(7x+1\right)\left(x+1\right)$$

To find equation solutions, solve 7x+1=0 and x+1=0.

$$x=-\frac{1}{7} x=-1$$

Steps Using the Quadratic Formula

Reorder the terms.

$$7+8\times \frac{1}{x}+x^{-2}=0$$

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

$$x\times 7+8\times 1+xx^{-2}=0$$

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

$$x\times 7+8\times 1+x^{-1}=0$$

Multiply 8 and 1 to get 8.

$$x\times 7+8+x^{-1}=0$$

Reorder the terms.

$$7x+8+\frac{1}{x}=0$$

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

$$7xx+x\times 8+1=0$$

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

$$7x^{2}+x\times 8+1=0$$

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.

$$7x^{2}+8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 7}}{2\times 7}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 7}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-8±\sqrt{64-28}}{2\times 7}$$

Add 64 to -28.

$$x=\frac{-8±\sqrt{36}}{2\times 7}$$

Take the square root of 36.

$$x=\frac{-8±6}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-8±6}{14}$$

Now solve the equation x=\frac{-8±6}{14} when ± is plus. Add -8 to 6.

$$x=-\frac{2}{14}$$

Reduce the fraction \frac{-2}{14} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{7}$$

Now solve the equation x=\frac{-8±6}{14} when ± is minus. Subtract 6 from -8.

$$x=-\frac{14}{14}$$

Divide -14 by 14.

$$x=-1$$

The equation is now solved.

$$x=-\frac{1}{7} x=-1$$

Steps for Completing the Square

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$x^{-2}+8x^{-1}=-7$$

Reorder the terms.

$$8\times \frac{1}{x}+x^{-2}=-7$$

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

$$8\times 1+xx^{-2}=-7x$$

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

$$8\times 1+x^{-1}=-7x$$

Multiply 8 and 1 to get 8.

$$8+x^{-1}=-7x$$

Add 7x to both sides.

$$8+x^{-1}+7x=0$$

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

$$x^{-1}+7x=-8$$

Reorder the terms.

$$7x+\frac{1}{x}=-8$$

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

$$7xx+1=-8x$$

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

$$7x^{2}+1=-8x$$

Add 8x to both sides.

$$7x^{2}+1+8x=0$$

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

$$7x^{2}+8x=-1$$

Divide both sides by 7.

$$\frac{7x^{2}+8x}{7}=-\frac{1}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{8}{7}x=-\frac{1}{7}$$

Divide \frac{8}{7}, the coefficient of the x term, by 2 to get \frac{4}{7}. Then add the square of \frac{4}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{7}x+\left(\frac{4}{7}\right)^{2}=-\frac{1}{7}+\left(\frac{4}{7}\right)^{2}$$

Square \frac{4}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=-\frac{1}{7}+\frac{16}{49}$$

Add -\frac{1}{7} to \frac{16}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=\frac{9}{49}$$

Factor x^{2}+\frac{8}{7}x+\frac{16}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{7}\right)^{2}=\frac{9}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{7}\right)^{2}}=\sqrt{\frac{9}{49}}$$

Simplify.

$$x+\frac{4}{7}=\frac{3}{7} x+\frac{4}{7}=-\frac{3}{7}$$

Subtract \frac{4}{7} from both sides of the equation.

$$x=-\frac{1}{7} x=-1$$