Solve 5x/x+2-20/x^2+2x=4/x

$$\frac { 5 x } { x + 2 } - \frac { 20 } { x ^ { 2 } + 2 x } = \frac { 4 } { x }$$

Solve for x

$$x = \frac{14}{5} = 2\frac{4}{5} = 2.8$$

Steps Using Factoring By Grouping

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

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

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

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

Use the distributive property to multiply x+2 by 4.

$$x^{2}\times 5-20=4x+8$$

Subtract 4x from both sides.

$$x^{2}\times 5-20-4x=8$$

Subtract 8 from both sides.

$$x^{2}\times 5-20-4x-8=0$$

Subtract 8 from -20 to get -28.

$$x^{2}\times 5-28-4x=0$$

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

$$5x^{2}-4x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=-4 ab=5\left(-28\right)=-140$$

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 -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -4.

$$a=-14 b=10$$

Rewrite 5x^{2}-4x-28 as \left(5x^{2}-14x\right)+\left(10x-28\right).

$$\left(5x^{2}-14x\right)+\left(10x-28\right)$$

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

$$x\left(5x-14\right)+2\left(5x-14\right)$$

Factor out common term 5x-14 by using distributive property.

$$\left(5x-14\right)\left(x+2\right)$$

To find equation solutions, solve 5x-14=0 and x+2=0.

$$x=\frac{14}{5} x=-2$$

Variable x cannot be equal to -2.

$$x=\frac{14}{5}$$

Steps Using the Quadratic Formula

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

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

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

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

Use the distributive property to multiply x+2 by 4.

$$x^{2}\times 5-20=4x+8$$

Subtract 4x from both sides.

$$x^{2}\times 5-20-4x=8$$

Subtract 8 from both sides.

$$x^{2}\times 5-20-4x-8=0$$

Subtract 8 from -20 to get -28.

$$x^{2}\times 5-28-4x=0$$

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.

$$5x^{2}-4x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\left(-28\right)}}{2\times 5}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-28\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\left(-28\right)}}{2\times 5}$$

Multiply -20 times -28.

$$x=\frac{-\left(-4\right)±\sqrt{16+560}}{2\times 5}$$

Add 16 to 560.

$$x=\frac{-\left(-4\right)±\sqrt{576}}{2\times 5}$$

Take the square root of 576.

$$x=\frac{-\left(-4\right)±24}{2\times 5}$$

The opposite of -4 is 4.

$$x=\frac{4±24}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{4±24}{10}$$

Now solve the equation x=\frac{4±24}{10} when ± is plus. Add 4 to 24.

$$x=\frac{28}{10}$$

Reduce the fraction \frac{28}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{14}{5}$$

Now solve the equation x=\frac{4±24}{10} when ± is minus. Subtract 24 from 4.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

The equation is now solved.

$$x=\frac{14}{5} x=-2$$

Variable x cannot be equal to -2.

$$x=\frac{14}{5}$$

Steps for Completing the Square

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

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

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

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

Use the distributive property to multiply x+2 by 4.

$$x^{2}\times 5-20=4x+8$$

Subtract 4x from both sides.

$$x^{2}\times 5-20-4x=8$$

Add 20 to both sides.

$$x^{2}\times 5-4x=8+20$$

Add 8 and 20 to get 28.

$$x^{2}\times 5-4x=28$$

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.

$$5x^{2}-4x=28$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{28}{5}+\left(-\frac{2}{5}\right)^{2}$$

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

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{28}{5}+\frac{4}{25}$$

Add \frac{28}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{144}{25}$$

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

$$\left(x-\frac{2}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{12}{5} x-\frac{2}{5}=-\frac{12}{5}$$

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

$$x=\frac{14}{5} x=-2$$

Variable x cannot be equal to -2.

$$x=\frac{14}{5}$$