Solve 1/x^2+8/x+15=0

$$\frac { 1 } { x ^ { 2 } } + \frac { 8 } { x } + 15 = 0$$

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

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

$$1+x\times 8+x^{2}\times 15=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$15x^{2}+8x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=8 ab=15\times 1=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 8.

$$a=3 b=5$$

Rewrite 15x^{2}+8x+1 as \left(15x^{2}+3x\right)+\left(5x+1\right).

$$\left(15x^{2}+3x\right)+\left(5x+1\right)$$

Factor out 3x in 15x^{2}+3x.

$$3x\left(5x+1\right)+5x+1$$

Factor out common term 5x+1 by using distributive property.

$$\left(5x+1\right)\left(3x+1\right)$$

To find equation solutions, solve 5x+1=0 and 3x+1=0.

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

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}, the least common multiple of x^{2},x.

$$1+x\times 8+x^{2}\times 15=0$$

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.

$$15x^{2}+8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 15}}{2\times 15}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 15}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-8±\sqrt{64-60}}{2\times 15}$$

Add 64 to -60.

$$x=\frac{-8±\sqrt{4}}{2\times 15}$$

Take the square root of 4.

$$x=\frac{-8±2}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-8±2}{30}$$

Now solve the equation x=\frac{-8±2}{30} when ± is plus. Add -8 to 2.

$$x=-\frac{6}{30}$$

Reduce the fraction \frac{-6}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{5}$$

Now solve the equation x=\frac{-8±2}{30} when ± is minus. Subtract 2 from -8.

$$x=-\frac{10}{30}$$

Reduce the fraction \frac{-10}{30} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$1+x\times 8+x^{2}\times 15=0$$

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

$$x\times 8+x^{2}\times 15=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$15x^{2}+8x=-1$$

Divide both sides by 15.

$$\frac{15x^{2}+8x}{15}=-\frac{1}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\frac{8}{15}x=-\frac{1}{15}$$

Divide \frac{8}{15}, the coefficient of the x term, by 2 to get \frac{4}{15}. Then add the square of \frac{4}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{15}x+\left(\frac{4}{15}\right)^{2}=-\frac{1}{15}+\left(\frac{4}{15}\right)^{2}$$

Square \frac{4}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{15}x+\frac{16}{225}=-\frac{1}{15}+\frac{16}{225}$$

Add -\frac{1}{15} to \frac{16}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{15}x+\frac{16}{225}=\frac{1}{225}$$

Factor x^{2}+\frac{8}{15}x+\frac{16}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{15}\right)^{2}=\frac{1}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{15}\right)^{2}}=\sqrt{\frac{1}{225}}$$

Simplify.

$$x+\frac{4}{15}=\frac{1}{15} x+\frac{4}{15}=-\frac{1}{15}$$

Subtract \frac{4}{15} from both sides of the equation.

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