Solve -4/x^2+4x-1/x+4=-5

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

Solve for x

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

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x^{2}+4x,x+4.

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

Use the distributive property to multiply -5x by x+4.

$$-4-x=-5x^{2}-20x$$

Add 5x^{2} to both sides.

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

Add 20x to both sides.

$$-4-x+5x^{2}+20x=0$$

Combine -x and 20x to get 19x.

$$-4+19x+5x^{2}=0$$

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

$$5x^{2}+19x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=19 ab=5\left(-4\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 19.

$$a=-1 b=20$$

Rewrite 5x^{2}+19x-4 as \left(5x^{2}-x\right)+\left(20x-4\right).

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

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

$$x\left(5x-1\right)+4\left(5x-1\right)$$

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

$$\left(5x-1\right)\left(x+4\right)$$

To find equation solutions, solve 5x-1=0 and x+4=0.

$$x=\frac{1}{5} x=-4$$

Variable x cannot be equal to -4.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x^{2}+4x,x+4.

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

Use the distributive property to multiply -5x by x+4.

$$-4-x=-5x^{2}-20x$$

Add 5x^{2} to both sides.

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

Add 20x to both sides.

$$-4-x+5x^{2}+20x=0$$

Combine -x and 20x to get 19x.

$$-4+19x+5x^{2}=0$$

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.

$$5x^{2}+19x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 19 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-19±\sqrt{19^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-19±\sqrt{361-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$x=\frac{-19±\sqrt{361+80}}{2\times 5}$$

Add 361 to 80.

$$x=\frac{-19±\sqrt{441}}{2\times 5}$$

Take the square root of 441.

$$x=\frac{-19±21}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-19±21}{10}$$

Now solve the equation x=\frac{-19±21}{10} when ± is plus. Add -19 to 21.

$$x=\frac{2}{10}$$

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

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

Now solve the equation x=\frac{-19±21}{10} when ± is minus. Subtract 21 from -19.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

The equation is now solved.

$$x=\frac{1}{5} x=-4$$

Variable x cannot be equal to -4.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x^{2}+4x,x+4.

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

Use the distributive property to multiply -5x by x+4.

$$-4-x=-5x^{2}-20x$$

Add 5x^{2} to both sides.

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

Add 20x to both sides.

$$-4-x+5x^{2}+20x=0$$

Add 4 to both sides. Anything plus zero gives itself.

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

Combine -x and 20x to get 19x.

$$19x+5x^{2}=4$$

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.

$$5x^{2}+19x=4$$

Divide both sides by 5.

$$\frac{5x^{2}+19x}{5}=\frac{4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{19}{5}x=\frac{4}{5}$$

Divide \frac{19}{5}, the coefficient of the x term, by 2 to get \frac{19}{10}. Then add the square of \frac{19}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{5}x+\left(\frac{19}{10}\right)^{2}=\frac{4}{5}+\left(\frac{19}{10}\right)^{2}$$

Square \frac{19}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{5}x+\frac{361}{100}=\frac{4}{5}+\frac{361}{100}$$

Add \frac{4}{5} to \frac{361}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{5}x+\frac{361}{100}=\frac{441}{100}$$

Factor x^{2}+\frac{19}{5}x+\frac{361}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{10}\right)^{2}=\frac{441}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{10}\right)^{2}}=\sqrt{\frac{441}{100}}$$

Simplify.

$$x+\frac{19}{10}=\frac{21}{10} x+\frac{19}{10}=-\frac{21}{10}$$

Subtract \frac{19}{10} from both sides of the equation.

$$x=\frac{1}{5} x=-4$$

Variable x cannot be equal to -4.

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