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

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

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring

Subtract \frac{1}{2} from both sides.

$$\frac{-x}{-x^{2}-x+2}-\frac{1}{2}=0$$

Factor -x^{2}-x+2.

$$\frac{-x}{\left(x+2\right)\left(-x+1\right)}-\frac{1}{2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)\left(-x+1\right) and 2 is 2\left(x+2\right)\left(-x+1\right). Multiply \frac{-x}{\left(x+2\right)\left(-x+1\right)} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(x+2\right)\left(-x+1\right)}{\left(x+2\right)\left(-x+1\right)}.

$$\frac{2\left(-x\right)}{2\left(x+2\right)\left(-x+1\right)}-\frac{\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{2\left(-x\right)-\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{-2x+x^{2}-x+2x-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

Combine like terms in -2x+x^{2}-x+2x-2.

$$\frac{-x+x^{2}-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

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

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

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

To solve the equation, factor x^{2}-x-2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2$$

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.

$$a=-2 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Steps Using Factoring By Grouping

Subtract \frac{1}{2} from both sides.

$$\frac{-x}{-x^{2}-x+2}-\frac{1}{2}=0$$

Factor -x^{2}-x+2.

$$\frac{-x}{\left(x+2\right)\left(-x+1\right)}-\frac{1}{2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)\left(-x+1\right) and 2 is 2\left(x+2\right)\left(-x+1\right). Multiply \frac{-x}{\left(x+2\right)\left(-x+1\right)} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(x+2\right)\left(-x+1\right)}{\left(x+2\right)\left(-x+1\right)}.

$$\frac{2\left(-x\right)}{2\left(x+2\right)\left(-x+1\right)}-\frac{\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{2\left(-x\right)-\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{-2x+x^{2}-x+2x-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

Combine like terms in -2x+x^{2}-x+2x-2.

$$\frac{-x+x^{2}-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

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

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

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

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

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.

$$a=-2 b=1$$

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

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

Factor out x in x^{2}-2x.

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

Factor out common term x-2 by using distributive property.

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

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Subtract \frac{1}{2} from both sides.

$$\frac{-x}{-x^{2}-x+2}-\frac{1}{2}=0$$

Factor -x^{2}-x+2.

$$\frac{-x}{\left(x+2\right)\left(-x+1\right)}-\frac{1}{2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+2\right)\left(-x+1\right) and 2 is 2\left(x+2\right)\left(-x+1\right). Multiply \frac{-x}{\left(x+2\right)\left(-x+1\right)} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(x+2\right)\left(-x+1\right)}{\left(x+2\right)\left(-x+1\right)}.

$$\frac{2\left(-x\right)}{2\left(x+2\right)\left(-x+1\right)}-\frac{\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{2\left(-x\right)-\left(x+2\right)\left(-x+1\right)}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

$$\frac{-2x+x^{2}-x+2x-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

Combine like terms in -2x+x^{2}-x+2x-2.

$$\frac{-x+x^{2}-2}{2\left(x+2\right)\left(-x+1\right)}=0$$

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

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

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.

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

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

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

Multiply -4 times -2.

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

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2}$$

Now solve the equation x=\frac{1±3}{2} when ± is plus. Add 1 to 3.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=2 x=-1$$

Steps for Completing the Square

Reorder the terms.

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

Divide both sides by -1.

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

Express \frac{\frac{1}{2}}{-1} as a single fraction.

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

Multiply 2 and -1 to get -2.

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

Fraction \frac{1}{-2} can be rewritten as -\frac{1}{2} by extracting the negative sign.

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

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

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

Use the distributive property to multiply -1 by x-1.

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

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

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

Add x^{2} to both sides.

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

Add x to both sides.

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

Combine -2x and x to get -x.

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

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.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$