Solve 1/2xleft(x-3right)=3x

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

Solve for x

$$x=9$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply \frac{1}{2}x by x-3.

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

Multiply x and x to get x^{2}.

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

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

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

Subtract 3x from both sides.

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

Combine -\frac{3}{2}x and -3x to get -\frac{9}{2}x.

$$\frac{1}{2}x^{2}-\frac{9}{2}x=0$$

Factor out x.

$$x\left(\frac{1}{2}x-\frac{9}{2}\right)=0$$

To find equation solutions, solve x=0 and \frac{x-9}{2}=0.

$$x=0 x=9$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}x by x-3.

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

Multiply x and x to get x^{2}.

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

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

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

Subtract 3x from both sides.

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

Combine -\frac{3}{2}x and -3x to get -\frac{9}{2}x.

$$\frac{1}{2}x^{2}-\frac{9}{2}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -\frac{9}{2} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-\frac{9}{2}\right)^{2}.

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

The opposite of -\frac{9}{2} is \frac{9}{2}.

$$x=\frac{\frac{9}{2}±\frac{9}{2}}{2\times \frac{1}{2}}$$

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

$$x=\frac{\frac{9}{2}±\frac{9}{2}}{1}$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{9}{2}}{1} when ± is plus. Add \frac{9}{2} to \frac{9}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{9}{1}$$

Divide 9 by 1.

$$x=9$$

Now solve the equation x=\frac{\frac{9}{2}±\frac{9}{2}}{1} when ± is minus. Subtract \frac{9}{2} from \frac{9}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by 1.

$$x=0$$

The equation is now solved.

$$x=9 x=0$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}x by x-3.

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

Multiply x and x to get x^{2}.

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

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

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

Fraction \frac{-3}{2} can be rewritten as -\frac{3}{2} by extracting the negative sign.

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

Subtract 3x from both sides.

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

Combine -\frac{3}{2}x and -3x to get -\frac{9}{2}x.

$$\frac{1}{2}x^{2}-\frac{9}{2}x=0$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-\frac{9}{2}x}{\frac{1}{2}}=\frac{0}{\frac{1}{2}}$$

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

$$x^{2}+\left(-\frac{\frac{9}{2}}{\frac{1}{2}}\right)x=\frac{0}{\frac{1}{2}}$$

Divide -\frac{9}{2} by \frac{1}{2} by multiplying -\frac{9}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}-9x=\frac{0}{\frac{1}{2}}$$

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

$$x^{2}-9x=0$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=\frac{81}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{9}{2} x-\frac{9}{2}=-\frac{9}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=9 x=0$$