Solve 40x-1/2x^2=20x-1/6x^2

$$40x- \frac{ 1 }{ 2 } { x }^{ 2 } =20x- \frac{ 1 }{ 6 } { x }^{ 2 }$$

Solve for x

$$x=60$$

$$x=0$$

Steps Using Factoring

Subtract 20x from both sides.

$$40x-\frac{1}{2}x^{2}-20x=-\frac{1}{6}x^{2}$$

Combine 40x and -20x to get 20x.

$$20x-\frac{1}{2}x^{2}=-\frac{1}{6}x^{2}$$

Add \frac{1}{6}x^{2} to both sides.

$$20x-\frac{1}{2}x^{2}+\frac{1}{6}x^{2}=0$$

Combine -\frac{1}{2}x^{2} and \frac{1}{6}x^{2} to get -\frac{1}{3}x^{2}.

$$20x-\frac{1}{3}x^{2}=0$$

Factor out x.

$$x\left(20-\frac{1}{3}x\right)=0$$

To find equation solutions, solve x=0 and 20-\frac{x}{3}=0.

$$x=0 x=60$$

Steps Using the Quadratic Formula

Subtract 20x from both sides.

$$40x-\frac{1}{2}x^{2}-20x=-\frac{1}{6}x^{2}$$

Combine 40x and -20x to get 20x.

$$20x-\frac{1}{2}x^{2}=-\frac{1}{6}x^{2}$$

Add \frac{1}{6}x^{2} to both sides.

$$20x-\frac{1}{2}x^{2}+\frac{1}{6}x^{2}=0$$

Combine -\frac{1}{2}x^{2} and \frac{1}{6}x^{2} to get -\frac{1}{3}x^{2}.

$$20x-\frac{1}{3}x^{2}=0$$

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.

$$-\frac{1}{3}x^{2}+20x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of 20^{2}.

$$x=\frac{-20±20}{2\left(-\frac{1}{3}\right)}$$

Multiply 2 times -\frac{1}{3}.

$$x=\frac{-20±20}{-\frac{2}{3}}$$

Now solve the equation x=\frac{-20±20}{-\frac{2}{3}} when ± is plus. Add -20 to 20.

$$x=\frac{0}{-\frac{2}{3}}$$

Divide 0 by -\frac{2}{3} by multiplying 0 by the reciprocal of -\frac{2}{3}.

$$x=0$$

Now solve the equation x=\frac{-20±20}{-\frac{2}{3}} when ± is minus. Subtract 20 from -20.

$$x=-\frac{40}{-\frac{2}{3}}$$

Divide -40 by -\frac{2}{3} by multiplying -40 by the reciprocal of -\frac{2}{3}.

$$x=60$$

The equation is now solved.

$$x=0 x=60$$

Steps for Completing the Square

Subtract 20x from both sides.

$$40x-\frac{1}{2}x^{2}-20x=-\frac{1}{6}x^{2}$$

Combine 40x and -20x to get 20x.

$$20x-\frac{1}{2}x^{2}=-\frac{1}{6}x^{2}$$

Add \frac{1}{6}x^{2} to both sides.

$$20x-\frac{1}{2}x^{2}+\frac{1}{6}x^{2}=0$$

Combine -\frac{1}{2}x^{2} and \frac{1}{6}x^{2} to get -\frac{1}{3}x^{2}.

$$20x-\frac{1}{3}x^{2}=0$$

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.

$$-\frac{1}{3}x^{2}+20x=0$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}x^{2}+20x}{-\frac{1}{3}}=\frac{0}{-\frac{1}{3}}$$

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

$$x^{2}+\frac{20}{-\frac{1}{3}}x=\frac{0}{-\frac{1}{3}}$$

Divide 20 by -\frac{1}{3} by multiplying 20 by the reciprocal of -\frac{1}{3}.

$$x^{2}-60x=\frac{0}{-\frac{1}{3}}$$

Divide 0 by -\frac{1}{3} by multiplying 0 by the reciprocal of -\frac{1}{3}.

$$x^{2}-60x=0$$

Divide -60, the coefficient of the x term, by 2 to get -30. Then add the square of -30 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-60x+\left(-30\right)^{2}=\left(-30\right)^{2}$$

Square -30.

$$x^{2}-60x+900=900$$

Factor x^{2}-60x+900. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-30\right)^{2}=900$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-30\right)^{2}}=\sqrt{900}$$

Simplify.

$$x-30=30 x-30=-30$$

Add 30 to both sides of the equation.

$$x=60 x=0$$