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

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

Solve for x

$$x=-1$$

$$x=4$$

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 \left(x+2\right)\left(x-2\right)^{2}, the least common multiple of x^{2}-4x+4,x^{2}-4.

$$\left(x+2\right)\left(5-x\right)=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x+2 by 5-x and combine like terms.

$$3x-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$3x-x^{2}+10=x^{2}-3x+2$$

Subtract x^{2} from both sides.

$$3x-x^{2}+10-x^{2}=-3x+2$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3x-2x^{2}+10=-3x+2$$

Add 3x to both sides.

$$3x-2x^{2}+10+3x=2$$

Combine 3x and 3x to get 6x.

$$6x-2x^{2}+10=2$$

Subtract 2 from both sides.

$$6x-2x^{2}+10-2=0$$

Subtract 2 from 10 to get 8.

$$6x-2x^{2}+8=0$$

Divide both sides by 2.

$$3x-x^{2}+4=0$$

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

$$-x^{2}+3x+4=0$$

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

$$a+b=3 ab=-4=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).

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

Factor out -x in the first and -1 in the second group.

$$-x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

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

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

$$x=4 x=-1$$

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 \left(x+2\right)\left(x-2\right)^{2}, the least common multiple of x^{2}-4x+4,x^{2}-4.

$$\left(x+2\right)\left(5-x\right)=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x+2 by 5-x and combine like terms.

$$3x-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$3x-x^{2}+10=x^{2}-3x+2$$

Subtract x^{2} from both sides.

$$3x-x^{2}+10-x^{2}=-3x+2$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3x-2x^{2}+10=-3x+2$$

Add 3x to both sides.

$$3x-2x^{2}+10+3x=2$$

Combine 3x and 3x to get 6x.

$$6x-2x^{2}+10=2$$

Subtract 2 from both sides.

$$6x-2x^{2}+10-2=0$$

Subtract 2 from 10 to get 8.

$$6x-2x^{2}+8=0$$

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.

$$-2x^{2}+6x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{-6±\sqrt{36+64}}{2\left(-2\right)}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±10}{-4}$$

Now solve the equation x=\frac{-6±10}{-4} when ± is plus. Add -6 to 10.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-6±10}{-4} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=-1 x=4$$

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 \left(x+2\right)\left(x-2\right)^{2}, the least common multiple of x^{2}-4x+4,x^{2}-4.

$$\left(x+2\right)\left(5-x\right)=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x+2 by 5-x and combine like terms.

$$3x-x^{2}+10=\left(x-2\right)\left(x-1\right)$$

Use the distributive property to multiply x-2 by x-1 and combine like terms.

$$3x-x^{2}+10=x^{2}-3x+2$$

Subtract x^{2} from both sides.

$$3x-x^{2}+10-x^{2}=-3x+2$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$3x-2x^{2}+10=-3x+2$$

Add 3x to both sides.

$$3x-2x^{2}+10+3x=2$$

Combine 3x and 3x to get 6x.

$$6x-2x^{2}+10=2$$

Subtract 10 from both sides.

$$6x-2x^{2}=2-10$$

Subtract 10 from 2 to get -8.

$$6x-2x^{2}=-8$$

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.

$$-2x^{2}+6x=-8$$

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=-\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=-\frac{8}{-2}$$

Divide 6 by -2.

$$x^{2}-3x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x^{2}-3x=4$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=4 x=-1$$