Solve x+5/x-2-28/x^2-4=12/5

$$\frac { x + 5 } { x - 2 } - \frac { 28 } { x ^ { 2 } - 4 } = \frac { 12 } { 5 }$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-2\right)\left(x+2\right), the least common multiple of x-2,x^{2}-4,5.

$$\left(5x+10\right)\left(x+5\right)-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 5x+10 by x+5 and combine like terms.

$$5x^{2}+35x+50-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Multiply -5 and 28 to get -140.

$$5x^{2}+35x+50-140=12\left(x-2\right)\left(x+2\right)$$

Subtract 140 from 50 to get -90.

$$5x^{2}+35x-90=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 12 by x-2.

$$5x^{2}+35x-90=\left(12x-24\right)\left(x+2\right)$$

Use the distributive property to multiply 12x-24 by x+2 and combine like terms.

$$5x^{2}+35x-90=12x^{2}-48$$

Subtract 12x^{2} from both sides.

$$5x^{2}+35x-90-12x^{2}=-48$$

Combine 5x^{2} and -12x^{2} to get -7x^{2}.

$$-7x^{2}+35x-90=-48$$

Add 48 to both sides.

$$-7x^{2}+35x-90+48=0$$

Add -90 and 48 to get -42.

$$-7x^{2}+35x-42=0$$

Divide both sides by 7.

$$-x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -x^{2}+5x-6 as \left(-x^{2}+3x\right)+\left(2x-6\right).

$$\left(-x^{2}+3x\right)+\left(2x-6\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-3\right)+2\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+2\right)$$

To find equation solutions, solve x-3=0 and -x+2=0.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-2\right)\left(x+2\right), the least common multiple of x-2,x^{2}-4,5.

$$\left(5x+10\right)\left(x+5\right)-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 5x+10 by x+5 and combine like terms.

$$5x^{2}+35x+50-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Multiply -5 and 28 to get -140.

$$5x^{2}+35x+50-140=12\left(x-2\right)\left(x+2\right)$$

Subtract 140 from 50 to get -90.

$$5x^{2}+35x-90=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 12 by x-2.

$$5x^{2}+35x-90=\left(12x-24\right)\left(x+2\right)$$

Use the distributive property to multiply 12x-24 by x+2 and combine like terms.

$$5x^{2}+35x-90=12x^{2}-48$$

Subtract 12x^{2} from both sides.

$$5x^{2}+35x-90-12x^{2}=-48$$

Combine 5x^{2} and -12x^{2} to get -7x^{2}.

$$-7x^{2}+35x-90=-48$$

Add 48 to both sides.

$$-7x^{2}+35x-90+48=0$$

Add -90 and 48 to get -42.

$$-7x^{2}+35x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 35 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\left(-7\right)\left(-42\right)}}{2\left(-7\right)}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\left(-7\right)\left(-42\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-35±\sqrt{1225+28\left(-42\right)}}{2\left(-7\right)}$$

Multiply 28 times -42.

$$x=\frac{-35±\sqrt{1225-1176}}{2\left(-7\right)}$$

Add 1225 to -1176.

$$x=\frac{-35±\sqrt{49}}{2\left(-7\right)}$$

Take the square root of 49.

$$x=\frac{-35±7}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-35±7}{-14}$$

Now solve the equation x=\frac{-35±7}{-14} when ± is plus. Add -35 to 7.

$$x=-\frac{28}{-14}$$

Divide -28 by -14.

$$x=2$$

Now solve the equation x=\frac{-35±7}{-14} when ± is minus. Subtract 7 from -35.

$$x=-\frac{42}{-14}$$

Divide -42 by -14.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Variable x cannot be equal to 2.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by 5\left(x-2\right)\left(x+2\right), the least common multiple of x-2,x^{2}-4,5.

$$\left(5x+10\right)\left(x+5\right)-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 5x+10 by x+5 and combine like terms.

$$5x^{2}+35x+50-5\times 28=12\left(x-2\right)\left(x+2\right)$$

Multiply -5 and 28 to get -140.

$$5x^{2}+35x+50-140=12\left(x-2\right)\left(x+2\right)$$

Subtract 140 from 50 to get -90.

$$5x^{2}+35x-90=12\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 12 by x-2.

$$5x^{2}+35x-90=\left(12x-24\right)\left(x+2\right)$$

Use the distributive property to multiply 12x-24 by x+2 and combine like terms.

$$5x^{2}+35x-90=12x^{2}-48$$

Subtract 12x^{2} from both sides.

$$5x^{2}+35x-90-12x^{2}=-48$$

Combine 5x^{2} and -12x^{2} to get -7x^{2}.

$$-7x^{2}+35x-90=-48$$

Add 90 to both sides.

$$-7x^{2}+35x=-48+90$$

Add -48 and 90 to get 42.

$$-7x^{2}+35x=42$$

Divide both sides by -7.

$$\frac{-7x^{2}+35x}{-7}=\frac{42}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{35}{-7}x=\frac{42}{-7}$$

Divide 35 by -7.

$$x^{2}-5x=\frac{42}{-7}$$

Divide 42 by -7.

$$x^{2}-5x=-6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$