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

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

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

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

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

Add \frac{3}{2}x to both sides.

$$-x-\frac{1}{2}x^{2}+\frac{3}{2}x=0$$

Combine -x and \frac{3}{2}x to get \frac{1}{2}x.

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

Factor out x.

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

To find equation solutions, solve x=0 and \frac{1-x}{2}=0.

$$x=0 x=1$$

Steps Using the Quadratic Formula

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

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

Add \frac{3}{2}x to both sides.

$$-x-\frac{1}{2}x^{2}+\frac{3}{2}x=0$$

Combine -x and \frac{3}{2}x to get \frac{1}{2}x.

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

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.

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

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

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

Take the square root of \left(\frac{1}{2}\right)^{2}.

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

Multiply 2 times -\frac{1}{2}.

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

Now solve the equation x=\frac{-\frac{1}{2}±\frac{1}{2}}{-1} when ± is plus. Add -\frac{1}{2} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{-1}$$

Divide 0 by -1.

$$x=0$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{1}{2}}{-1} when ± is minus. Subtract \frac{1}{2} from -\frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -1 by -1.

$$x=1$$

The equation is now solved.

$$x=0 x=1$$

Steps for Completing the Square

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

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

Add \frac{3}{2}x to both sides.

$$-x-\frac{1}{2}x^{2}+\frac{3}{2}x=0$$

Combine -x and \frac{3}{2}x to get \frac{1}{2}x.

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

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.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide \frac{1}{2} by -\frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of -\frac{1}{2}.

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

Divide 0 by -\frac{1}{2} by multiplying 0 by the reciprocal of -\frac{1}{2}.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=1 x=0$$