Solve 1/2x(7-x)=6

$$\frac { 1 } { 2 } x ( 7 - x ) = 6$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}x by 7-x.

$$\frac{1}{2}x\times 7+\frac{1}{2}x\left(-1\right)x=6$$

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

$$\frac{1}{2}x\times 7+\frac{1}{2}x^{2}\left(-1\right)=6$$

Multiply \frac{1}{2} and 7 to get \frac{7}{2}.

$$\frac{7}{2}x+\frac{1}{2}x^{2}\left(-1\right)=6$$

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

$$\frac{7}{2}x-\frac{1}{2}x^{2}=6$$

Subtract 6 from both sides.

$$\frac{7}{2}x-\frac{1}{2}x^{2}-6=0$$

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.

$$-\frac{1}{2}x^{2}+\frac{7}{2}x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, \frac{7}{2} for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{7}{2}±\sqrt{\left(\frac{7}{2}\right)^{2}-4\left(-\frac{1}{2}\right)\left(-6\right)}}{2\left(-\frac{1}{2}\right)}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{49}{4}-4\left(-\frac{1}{2}\right)\left(-6\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{49}{4}+2\left(-6\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -6.

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{49}{4}-12}}{2\left(-\frac{1}{2}\right)}$$

Add \frac{49}{4} to -12.

$$x=\frac{-\frac{7}{2}±\sqrt{\frac{1}{4}}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of \frac{1}{4}.

$$x=\frac{-\frac{7}{2}±\frac{1}{2}}{2\left(-\frac{1}{2}\right)}$$

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

$$x=\frac{-\frac{7}{2}±\frac{1}{2}}{-1}$$

Now solve the equation x=\frac{-\frac{7}{2}±\frac{1}{2}}{-1} when ± is plus. Add -\frac{7}{2} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -3 by -1.

$$x=3$$

Now solve the equation x=\frac{-\frac{7}{2}±\frac{1}{2}}{-1} when ± is minus. Subtract \frac{1}{2} from -\frac{7}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x=4$$

The equation is now solved.

$$x=3 x=4$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}x by 7-x.

$$\frac{1}{2}x\times 7+\frac{1}{2}x\left(-1\right)x=6$$

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

$$\frac{1}{2}x\times 7+\frac{1}{2}x^{2}\left(-1\right)=6$$

Multiply \frac{1}{2} and 7 to get \frac{7}{2}.

$$\frac{7}{2}x+\frac{1}{2}x^{2}\left(-1\right)=6$$

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

$$\frac{7}{2}x-\frac{1}{2}x^{2}=6$$

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.

$$-\frac{1}{2}x^{2}+\frac{7}{2}x=6$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}+\frac{7}{2}x}{-\frac{1}{2}}=\frac{6}{-\frac{1}{2}}$$

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

$$x^{2}+\frac{\frac{7}{2}}{-\frac{1}{2}}x=\frac{6}{-\frac{1}{2}}$$

Divide \frac{7}{2} by -\frac{1}{2} by multiplying \frac{7}{2} by the reciprocal of -\frac{1}{2}.

$$x^{2}-7x=\frac{6}{-\frac{1}{2}}$$

Divide 6 by -\frac{1}{2} by multiplying 6 by the reciprocal of -\frac{1}{2}.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$