Solve x-2/x-1-x-1/x-2=x^2/x^2-3x+2

$$\frac { x - 2 } { x - 1 } - \frac { x - 1 } { x - 2 } = \frac { x ^ { 2 } } { x ^ { 2 } - 3 x + 2 }$$

Solve for x

$$x=-3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-1,x-2,x^{2}-3x+2.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$x^{2}-4x+4-x^{2}+2x-1=x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+4+2x-1=x^{2}$$

Combine -4x and 2x to get -2x.

$$-2x+4-1=x^{2}$$

Subtract 1 from 4 to get 3.

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

Subtract x^{2} from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

Factor out x in the first and 3 in the second group.

$$x\left(-x+1\right)+3\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+3\right)$$

To find equation solutions, solve -x+1=0 and x+3=0.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-1,x-2,x^{2}-3x+2.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$x^{2}-4x+4-x^{2}+2x-1=x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+4+2x-1=x^{2}$$

Combine -4x and 2x to get -2x.

$$-2x+4-1=x^{2}$$

Subtract 1 from 4 to get 3.

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

Subtract x^{2} from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-2\right)±\sqrt{4+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-1\right)}$$

Add 4 to 12.

$$x=\frac{-\left(-2\right)±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

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

The opposite of -2 is 2.

$$x=\frac{2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{2±4}{-2}$$

Now solve the equation x=\frac{2±4}{-2} when ± is plus. Add 2 to 4.

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{2±4}{-2} when ± is minus. Subtract 4 from 2.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-3 x=1$$

Variable x cannot be equal to 1.

$$x=-3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-1,x-2,x^{2}-3x+2.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$x^{2}-4x+4-x^{2}+2x-1=x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-4x+4+2x-1=x^{2}$$

Combine -4x and 2x to get -2x.

$$-2x+4-1=x^{2}$$

Subtract 1 from 4 to get 3.

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

Subtract x^{2} from both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

$$x^{2}+2x=-\frac{3}{-1}$$

Divide -3 by -1.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$

Variable x cannot be equal to 1.

$$x=-3$$