Solve 2/x^2+5x+3/2x-10=15/x^2-25

$$\frac { 2 } { x ^ { 2 } + 5 x } + \frac { 3 } { 2 x - 10 } = \frac { 15 } { x ^ { 2 } - 25 }$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,0,5 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}+5x,2x-10,x^{2}-25.

$$\left(2x-10\right)\times 2+x\left(x+5\right)\times 3=2x\times 15$$

Use the distributive property to multiply 2x-10 by 2.

$$4x-20+x\left(x+5\right)\times 3=2x\times 15$$

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

$$4x-20+\left(x^{2}+5x\right)\times 3=2x\times 15$$

Use the distributive property to multiply x^{2}+5x by 3.

$$4x-20+3x^{2}+15x=2x\times 15$$

Combine 4x and 15x to get 19x.

$$19x-20+3x^{2}=2x\times 15$$

Multiply 2 and 15 to get 30.

$$19x-20+3x^{2}=30x$$

Subtract 30x from both sides.

$$19x-20+3x^{2}-30x=0$$

Combine 19x and -30x to get -11x.

$$-11x-20+3x^{2}=0$$

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

$$3x^{2}-11x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\left(-20\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$a=-15 b=4$$

Rewrite 3x^{2}-11x-20 as \left(3x^{2}-15x\right)+\left(4x-20\right).

$$\left(3x^{2}-15x\right)+\left(4x-20\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(x-5\right)+4\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(3x+4\right)$$

To find equation solutions, solve x-5=0 and 3x+4=0.

$$x=5 x=-\frac{4}{3}$$

Variable x cannot be equal to 5.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,0,5 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}+5x,2x-10,x^{2}-25.

$$\left(2x-10\right)\times 2+x\left(x+5\right)\times 3=2x\times 15$$

Use the distributive property to multiply 2x-10 by 2.

$$4x-20+x\left(x+5\right)\times 3=2x\times 15$$

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

$$4x-20+\left(x^{2}+5x\right)\times 3=2x\times 15$$

Use the distributive property to multiply x^{2}+5x by 3.

$$4x-20+3x^{2}+15x=2x\times 15$$

Combine 4x and 15x to get 19x.

$$19x-20+3x^{2}=2x\times 15$$

Multiply 2 and 15 to get 30.

$$19x-20+3x^{2}=30x$$

Subtract 30x from both sides.

$$19x-20+3x^{2}-30x=0$$

Combine 19x and -30x to get -11x.

$$-11x-20+3x^{2}=0$$

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.

$$3x^{2}-11x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\left(-20\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\left(-20\right)}}{2\times 3}$$

Multiply -12 times -20.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 3}$$

Add 121 to 240.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\times 3}$$

Take the square root of 361.

$$x=\frac{-\left(-11\right)±19}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±19}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±19}{6}$$

Now solve the equation x=\frac{11±19}{6} when ± is plus. Add 11 to 19.

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

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{11±19}{6} when ± is minus. Subtract 19 from 11.

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

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

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

The equation is now solved.

$$x=5 x=-\frac{4}{3}$$

Variable x cannot be equal to 5.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,0,5 since division by zero is not defined. Multiply both sides of the equation by 2x\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}+5x,2x-10,x^{2}-25.

$$\left(2x-10\right)\times 2+x\left(x+5\right)\times 3=2x\times 15$$

Use the distributive property to multiply 2x-10 by 2.

$$4x-20+x\left(x+5\right)\times 3=2x\times 15$$

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

$$4x-20+\left(x^{2}+5x\right)\times 3=2x\times 15$$

Use the distributive property to multiply x^{2}+5x by 3.

$$4x-20+3x^{2}+15x=2x\times 15$$

Combine 4x and 15x to get 19x.

$$19x-20+3x^{2}=2x\times 15$$

Multiply 2 and 15 to get 30.

$$19x-20+3x^{2}=30x$$

Subtract 30x from both sides.

$$19x-20+3x^{2}-30x=0$$

Combine 19x and -30x to get -11x.

$$-11x-20+3x^{2}=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$-11x+3x^{2}=20$$

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.

$$3x^{2}-11x=20$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=\frac{20}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=\frac{20}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{20}{3}+\frac{121}{36}$$

Add \frac{20}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{361}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{361}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{19}{6} x-\frac{11}{6}=-\frac{19}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=5 x=-\frac{4}{3}$$

Variable x cannot be equal to 5.

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