Solve {left(-1/x-2right)}^2-2-2/x-2=5

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

Solve for x

$$x=3$$

$$x = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

Calculate -\frac{1}{x-2} to the power of 2 and get \left(\frac{1}{x-2}\right)^{2}.

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

Use the distributive property to multiply x-2 by \left(\frac{1}{x-2}\right)^{2}.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express x\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express -2\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

Since \frac{x\times 1^{2}}{\left(x-2\right)^{2}} and \frac{-2\times 1^{2}}{\left(x-2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x\times 1^{2}-2\times 1^{2}.

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

Rewrite \left(x-2\right)^{2} as \left(x-2\right)\left(x-2\right). Cancel out x-2 in both numerator and denominator.

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

Multiply 2 and -2 to get -4.

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

The opposite of -4 is 4.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{x-2}{x-2}.

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

Since \frac{1}{x-2} and \frac{4\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 1+4\left(x-2\right).

$$\frac{1+4x-8}{x-2}=5\left(x-2\right)$$

Combine like terms in 1+4x-8.

$$\frac{-7+4x}{x-2}=5\left(x-2\right)$$

Use the distributive property to multiply 5 by x-2.

$$\frac{-7+4x}{x-2}=5x-10$$

Subtract 5x from both sides.

$$\frac{-7+4x}{x-2}-5x=-10$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -5x times \frac{x-2}{x-2}.

$$\frac{-7+4x}{x-2}+\frac{-5x\left(x-2\right)}{x-2}=-10$$

Since \frac{-7+4x}{x-2} and \frac{-5x\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

$$\frac{-7+4x-5x\left(x-2\right)}{x-2}=-10$$

Do the multiplications in -7+4x-5x\left(x-2\right).

$$\frac{-7+4x-5x^{2}+10x}{x-2}=-10$$

Combine like terms in -7+4x-5x^{2}+10x.

$$\frac{-7+14x-5x^{2}}{x-2}=-10$$

Add 10 to both sides.

$$\frac{-7+14x-5x^{2}}{x-2}+10=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 10 times \frac{x-2}{x-2}.

$$\frac{-7+14x-5x^{2}}{x-2}+\frac{10\left(x-2\right)}{x-2}=0$$

Since \frac{-7+14x-5x^{2}}{x-2} and \frac{10\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

$$\frac{-7+14x-5x^{2}+10\left(x-2\right)}{x-2}=0$$

Do the multiplications in -7+14x-5x^{2}+10\left(x-2\right).

$$\frac{-7+14x-5x^{2}+10x-20}{x-2}=0$$

Combine like terms in -7+14x-5x^{2}+10x-20.

$$\frac{-27+24x-5x^{2}}{x-2}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$-27+24x-5x^{2}=0$$

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

$$-5x^{2}+24x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=24 ab=-5\left(-27\right)=135$$

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 135.

$$1,135 3,45 5,27 9,15$$

Calculate the sum for each pair.

$$1+135=136 3+45=48 5+27=32 9+15=24$$

The solution is the pair that gives sum 24.

$$a=15 b=9$$

Rewrite -5x^{2}+24x-27 as \left(-5x^{2}+15x\right)+\left(9x-27\right).

$$\left(-5x^{2}+15x\right)+\left(9x-27\right)$$

Factor out 5x in the first and -9 in the second group.

$$5x\left(-x+3\right)-9\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(5x-9\right)$$

To find equation solutions, solve -x+3=0 and 5x-9=0.

$$x=3 x=\frac{9}{5}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

Calculate -\frac{1}{x-2} to the power of 2 and get \left(\frac{1}{x-2}\right)^{2}.

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

Use the distributive property to multiply x-2 by \left(\frac{1}{x-2}\right)^{2}.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express x\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express -2\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

Since \frac{x\times 1^{2}}{\left(x-2\right)^{2}} and \frac{-2\times 1^{2}}{\left(x-2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x\times 1^{2}-2\times 1^{2}.

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

Rewrite \left(x-2\right)^{2} as \left(x-2\right)\left(x-2\right). Cancel out x-2 in both numerator and denominator.

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

Multiply 2 and -2 to get -4.

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

The opposite of -4 is 4.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{x-2}{x-2}.

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

Since \frac{1}{x-2} and \frac{4\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 1+4\left(x-2\right).

$$\frac{1+4x-8}{x-2}=5\left(x-2\right)$$

Combine like terms in 1+4x-8.

$$\frac{-7+4x}{x-2}=5\left(x-2\right)$$

Use the distributive property to multiply 5 by x-2.

$$\frac{-7+4x}{x-2}=5x-10$$

Subtract 5x from both sides.

$$\frac{-7+4x}{x-2}-5x=-10$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -5x times \frac{x-2}{x-2}.

$$\frac{-7+4x}{x-2}+\frac{-5x\left(x-2\right)}{x-2}=-10$$

Since \frac{-7+4x}{x-2} and \frac{-5x\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

$$\frac{-7+4x-5x\left(x-2\right)}{x-2}=-10$$

Do the multiplications in -7+4x-5x\left(x-2\right).

$$\frac{-7+4x-5x^{2}+10x}{x-2}=-10$$

Combine like terms in -7+4x-5x^{2}+10x.

$$\frac{-7+14x-5x^{2}}{x-2}=-10$$

Add 10 to both sides.

$$\frac{-7+14x-5x^{2}}{x-2}+10=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 10 times \frac{x-2}{x-2}.

$$\frac{-7+14x-5x^{2}}{x-2}+\frac{10\left(x-2\right)}{x-2}=0$$

Since \frac{-7+14x-5x^{2}}{x-2} and \frac{10\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

$$\frac{-7+14x-5x^{2}+10\left(x-2\right)}{x-2}=0$$

Do the multiplications in -7+14x-5x^{2}+10\left(x-2\right).

$$\frac{-7+14x-5x^{2}+10x-20}{x-2}=0$$

Combine like terms in -7+14x-5x^{2}+10x-20.

$$\frac{-27+24x-5x^{2}}{x-2}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$-27+24x-5x^{2}=0$$

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.

$$-5x^{2}+24x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 24 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\left(-5\right)\left(-27\right)}}{2\left(-5\right)}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\left(-5\right)\left(-27\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-24±\sqrt{576+20\left(-27\right)}}{2\left(-5\right)}$$

Multiply 20 times -27.

$$x=\frac{-24±\sqrt{576-540}}{2\left(-5\right)}$$

Add 576 to -540.

$$x=\frac{-24±\sqrt{36}}{2\left(-5\right)}$$

Take the square root of 36.

$$x=\frac{-24±6}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-24±6}{-10}$$

Now solve the equation x=\frac{-24±6}{-10} when ± is plus. Add -24 to 6.

$$x=-\frac{18}{-10}$$

Reduce the fraction \frac{-18}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{5}$$

Now solve the equation x=\frac{-24±6}{-10} when ± is minus. Subtract 6 from -24.

$$x=-\frac{30}{-10}$$

Divide -30 by -10.

$$x=3$$

The equation is now solved.

$$x=\frac{9}{5} x=3$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

Calculate -\frac{1}{x-2} to the power of 2 and get \left(\frac{1}{x-2}\right)^{2}.

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

Use the distributive property to multiply x-2 by \left(\frac{1}{x-2}\right)^{2}.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express x\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

To raise \frac{1}{x-2} to a power, raise both numerator and denominator to the power and then divide.

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

Express -2\times \frac{1^{2}}{\left(x-2\right)^{2}} as a single fraction.

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

Since \frac{x\times 1^{2}}{\left(x-2\right)^{2}} and \frac{-2\times 1^{2}}{\left(x-2\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x\times 1^{2}-2\times 1^{2}.

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

Rewrite \left(x-2\right)^{2} as \left(x-2\right)\left(x-2\right). Cancel out x-2 in both numerator and denominator.

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

Multiply 2 and -2 to get -4.

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

The opposite of -4 is 4.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{x-2}{x-2}.

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

Since \frac{1}{x-2} and \frac{4\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 1+4\left(x-2\right).

$$\frac{1+4x-8}{x-2}=5\left(x-2\right)$$

Combine like terms in 1+4x-8.

$$\frac{-7+4x}{x-2}=5\left(x-2\right)$$

Use the distributive property to multiply 5 by x-2.

$$\frac{-7+4x}{x-2}=5x-10$$

Subtract 5x from both sides.

$$\frac{-7+4x}{x-2}-5x=-10$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -5x times \frac{x-2}{x-2}.

$$\frac{-7+4x}{x-2}+\frac{-5x\left(x-2\right)}{x-2}=-10$$

Since \frac{-7+4x}{x-2} and \frac{-5x\left(x-2\right)}{x-2} have the same denominator, add them by adding their numerators.

$$\frac{-7+4x-5x\left(x-2\right)}{x-2}=-10$$

Do the multiplications in -7+4x-5x\left(x-2\right).

$$\frac{-7+4x-5x^{2}+10x}{x-2}=-10$$

Combine like terms in -7+4x-5x^{2}+10x.

$$\frac{-7+14x-5x^{2}}{x-2}=-10$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$-7+14x-5x^{2}=-10\left(x-2\right)$$

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

$$-7+14x-5x^{2}=-10x+20$$

Add 10x to both sides.

$$-7+14x-5x^{2}+10x=20$$

Combine 14x and 10x to get 24x.

$$-7+24x-5x^{2}=20$$

Add 7 to both sides.

$$24x-5x^{2}=20+7$$

Add 20 and 7 to get 27.

$$24x-5x^{2}=27$$

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.

$$-5x^{2}+24x=27$$

Divide both sides by -5.

$$\frac{-5x^{2}+24x}{-5}=\frac{27}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{24}{-5}x=\frac{27}{-5}$$

Divide 24 by -5.

$$x^{2}-\frac{24}{5}x=\frac{27}{-5}$$

Divide 27 by -5.

$$x^{2}-\frac{24}{5}x=-\frac{27}{5}$$

Divide -\frac{24}{5}, the coefficient of the x term, by 2 to get -\frac{12}{5}. Then add the square of -\frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{5}x+\left(-\frac{12}{5}\right)^{2}=-\frac{27}{5}+\left(-\frac{12}{5}\right)^{2}$$

Square -\frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=-\frac{27}{5}+\frac{144}{25}$$

Add -\frac{27}{5} to \frac{144}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=\frac{9}{25}$$

Factor x^{2}-\frac{24}{5}x+\frac{144}{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}.

$$\left(x-\frac{12}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x-\frac{12}{5}=\frac{3}{5} x-\frac{12}{5}=-\frac{3}{5}$$

Add \frac{12}{5} to both sides of the equation.

$$x=3 x=\frac{9}{5}$$