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

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

Solve for x

$$x = \frac{11}{5} = 2\frac{1}{5} = 2.2$$

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 \left(x-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Use the distributive property to multiply 5 by x^{2}-4x+4.

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

Subtract 5x^{2} from both sides.

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

Add 20x to both sides.

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

Combine x and 20x to get 21x.

$$21x-2-5x^{2}=20$$

Subtract 20 from both sides.

$$21x-2-5x^{2}-20=0$$

Subtract 20 from -2 to get -22.

$$21x-22-5x^{2}=0$$

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

$$-5x^{2}+21x-22=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx-22. To find a and b, set up a system to be solved.

$$a+b=21 ab=-5\left(-22\right)=110$$

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 110.

$$1,110 2,55 5,22 10,11$$

Calculate the sum for each pair.

$$1+110=111 2+55=57 5+22=27 10+11=21$$

The solution is the pair that gives sum 21.

$$a=11 b=10$$

Rewrite -5x^{2}+21x-22 as \left(-5x^{2}+11x\right)+\left(10x-22\right).

$$\left(-5x^{2}+11x\right)+\left(10x-22\right)$$

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

$$-x\left(5x-11\right)+2\left(5x-11\right)$$

Factor out common term 5x-11 by using distributive property.

$$\left(5x-11\right)\left(-x+2\right)$$

To find equation solutions, solve 5x-11=0 and -x+2=0.

$$x=\frac{11}{5} x=2$$

Variable x cannot be equal to 2.

$$x=\frac{11}{5}$$

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 \left(x-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Use the distributive property to multiply 5 by x^{2}-4x+4.

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

Subtract 5x^{2} from both sides.

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

Add 20x to both sides.

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

Combine x and 20x to get 21x.

$$21x-2-5x^{2}=20$$

Subtract 20 from both sides.

$$21x-2-5x^{2}-20=0$$

Subtract 20 from -2 to get -22.

$$21x-22-5x^{2}=0$$

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.

$$-5x^{2}+21x-22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 21 for b, and -22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-21±\sqrt{21^{2}-4\left(-5\right)\left(-22\right)}}{2\left(-5\right)}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\left(-5\right)\left(-22\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-21±\sqrt{441+20\left(-22\right)}}{2\left(-5\right)}$$

Multiply 20 times -22.

$$x=\frac{-21±\sqrt{441-440}}{2\left(-5\right)}$$

Add 441 to -440.

$$x=\frac{-21±\sqrt{1}}{2\left(-5\right)}$$

Take the square root of 1.

$$x=\frac{-21±1}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-21±1}{-10}$$

Now solve the equation x=\frac{-21±1}{-10} when ± is plus. Add -21 to 1.

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

Divide -20 by -10.

$$x=2$$

Now solve the equation x=\frac{-21±1}{-10} when ± is minus. Subtract 1 from -21.

$$x=-\frac{22}{-10}$$

Reduce the fraction \frac{-22}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{5}$$

The equation is now solved.

$$x=2 x=\frac{11}{5}$$

Variable x cannot be equal to 2.

$$x=\frac{11}{5}$$

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 \left(x-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Use the distributive property to multiply 5 by x^{2}-4x+4.

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

Subtract 5x^{2} from both sides.

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

Add 20x to both sides.

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

Combine x and 20x to get 21x.

$$21x-2-5x^{2}=20$$

Add 2 to both sides.

$$21x-5x^{2}=20+2$$

Add 20 and 2 to get 22.

$$21x-5x^{2}=22$$

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.

$$-5x^{2}+21x=22$$

Divide both sides by -5.

$$\frac{-5x^{2}+21x}{-5}=\frac{22}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{21}{-5}x=\frac{22}{-5}$$

Divide 21 by -5.

$$x^{2}-\frac{21}{5}x=\frac{22}{-5}$$

Divide 22 by -5.

$$x^{2}-\frac{21}{5}x=-\frac{22}{5}$$

Divide -\frac{21}{5}, the coefficient of the x term, by 2 to get -\frac{21}{10}. Then add the square of -\frac{21}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{21}{5}x+\left(-\frac{21}{10}\right)^{2}=-\frac{22}{5}+\left(-\frac{21}{10}\right)^{2}$$

Square -\frac{21}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{21}{5}x+\frac{441}{100}=-\frac{22}{5}+\frac{441}{100}$$

Add -\frac{22}{5} to \frac{441}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{21}{5}x+\frac{441}{100}=\frac{1}{100}$$

Factor x^{2}-\frac{21}{5}x+\frac{441}{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}.

$$\left(x-\frac{21}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{21}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$x-\frac{21}{10}=\frac{1}{10} x-\frac{21}{10}=-\frac{1}{10}$$

Add \frac{21}{10} to both sides of the equation.

$$x=\frac{11}{5} x=2$$

Variable x cannot be equal to 2.

$$x=\frac{11}{5}$$