Solve 3x/x-5+1/x+5=-10/x^2-25

$$\frac { 3 x } { x - 5 } + \frac { 1 } { x + 5 } = \frac { - 10 } { x ^ { 2 } - 25 }$$

Solve for x

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

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x-5,x+5,x^{2}-25.

$$\left(x+5\right)\times 3x+x-5=-10$$

Use the distributive property to multiply x+5 by 3.

$$\left(3x+15\right)x+x-5=-10$$

Use the distributive property to multiply 3x+15 by x.

$$3x^{2}+15x+x-5=-10$$

Combine 15x and x to get 16x.

$$3x^{2}+16x-5=-10$$

Add 10 to both sides.

$$3x^{2}+16x-5+10=0$$

Add -5 and 10 to get 5.

$$3x^{2}+16x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\times 5=15$$

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 16.

$$a=1 b=15$$

Rewrite 3x^{2}+16x+5 as \left(3x^{2}+x\right)+\left(15x+5\right).

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

Factor out x in the first and 5 in the second group.

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

Factor out common term 3x+1 by using distributive property.

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

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

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

Variable x cannot be equal to -5.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x-5,x+5,x^{2}-25.

$$\left(x+5\right)\times 3x+x-5=-10$$

Use the distributive property to multiply x+5 by 3.

$$\left(3x+15\right)x+x-5=-10$$

Use the distributive property to multiply 3x+15 by x.

$$3x^{2}+15x+x-5=-10$$

Combine 15x and x to get 16x.

$$3x^{2}+16x-5=-10$$

Add 10 to both sides.

$$3x^{2}+16x-5+10=0$$

Add -5 and 10 to get 5.

$$3x^{2}+16x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 16 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 3\times 5}}{2\times 3}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 3\times 5}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-16±\sqrt{256-12\times 5}}{2\times 3}$$

Multiply -12 times 5.

$$x=\frac{-16±\sqrt{256-60}}{2\times 3}$$

Add 256 to -60.

$$x=\frac{-16±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-16±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±14}{6}$$

Now solve the equation x=\frac{-16±14}{6} when ± is plus. Add -16 to 14.

$$x=-\frac{2}{6}$$

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

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

Now solve the equation x=\frac{-16±14}{6} when ± is minus. Subtract 14 from -16.

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

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

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

Variable x cannot be equal to -5.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x-5,x+5,x^{2}-25.

$$\left(x+5\right)\times 3x+x-5=-10$$

Use the distributive property to multiply x+5 by 3.

$$\left(3x+15\right)x+x-5=-10$$

Use the distributive property to multiply 3x+15 by x.

$$3x^{2}+15x+x-5=-10$$

Combine 15x and x to get 16x.

$$3x^{2}+16x-5=-10$$

Add 5 to both sides.

$$3x^{2}+16x=-10+5$$

Add -10 and 5 to get -5.

$$3x^{2}+16x=-5$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide \frac{16}{3}, the coefficient of the x term, by 2 to get \frac{8}{3}. Then add the square of \frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{16}{3}x+\left(\frac{8}{3}\right)^{2}=-\frac{5}{3}+\left(\frac{8}{3}\right)^{2}$$

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

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=-\frac{5}{3}+\frac{64}{9}$$

Add -\frac{5}{3} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=\frac{49}{9}$$

Factor x^{2}+\frac{16}{3}x+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{8}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{8}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x+\frac{8}{3}=\frac{7}{3} x+\frac{8}{3}=-\frac{7}{3}$$

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

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

Variable x cannot be equal to -5.

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