Solve (3x)(33-3x)=72

$$( 3 x ) ( 33 - 3 x ) = 72$$

Solve for x

$$x = \frac{\sqrt{89} + 11}{2} \approx 10.216990566$$

$$x=\frac{11-\sqrt{89}}{2}\approx 0.783009434$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 33-3x.

$$99x-9x^{2}=72$$

Subtract 72 from both sides.

$$99x-9x^{2}-72=0$$

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.

$$-9x^{2}+99x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 99 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-99±\sqrt{99^{2}-4\left(-9\right)\left(-72\right)}}{2\left(-9\right)}$$

Square 99.

$$x=\frac{-99±\sqrt{9801-4\left(-9\right)\left(-72\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-99±\sqrt{9801+36\left(-72\right)}}{2\left(-9\right)}$$

Multiply 36 times -72.

$$x=\frac{-99±\sqrt{9801-2592}}{2\left(-9\right)}$$

Add 9801 to -2592.

$$x=\frac{-99±\sqrt{7209}}{2\left(-9\right)}$$

Take the square root of 7209.

$$x=\frac{-99±9\sqrt{89}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-99±9\sqrt{89}}{-18}$$

Now solve the equation x=\frac{-99±9\sqrt{89}}{-18} when ± is plus. Add -99 to 9\sqrt{89}.

$$x=\frac{9\sqrt{89}-99}{-18}$$

Divide -99+9\sqrt{89} by -18.

$$x=\frac{11-\sqrt{89}}{2}$$

Now solve the equation x=\frac{-99±9\sqrt{89}}{-18} when ± is minus. Subtract 9\sqrt{89} from -99.

$$x=\frac{-9\sqrt{89}-99}{-18}$$

Divide -99-9\sqrt{89} by -18.

$$x=\frac{\sqrt{89}+11}{2}$$

The equation is now solved.

$$x=\frac{11-\sqrt{89}}{2} x=\frac{\sqrt{89}+11}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 33-3x.

$$99x-9x^{2}=72$$

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.

$$-9x^{2}+99x=72$$

Divide both sides by -9.

$$\frac{-9x^{2}+99x}{-9}=\frac{72}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{99}{-9}x=\frac{72}{-9}$$

Divide 99 by -9.

$$x^{2}-11x=\frac{72}{-9}$$

Divide 72 by -9.

$$x^{2}-11x=-8$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-8+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-8+\frac{121}{4}$$

Add -8 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{89}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{89}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{89}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{\sqrt{89}}{2} x-\frac{11}{2}=-\frac{\sqrt{89}}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=\frac{\sqrt{89}+11}{2} x=\frac{11-\sqrt{89}}{2}$$