Solve 9xleft(2-3xright)=1/2+3/8+4/6+5/2

$$9x \left( 2-3x \right) = \frac{ 1 }{ 2 } + \frac{ 3 }{ 8 } + \frac{ 4 }{ 6 } + \frac{ 5 }{ 2 }$$

Solve for x (complex solution)

$$x=\frac{5\sqrt{2}i}{36}+\frac{1}{3}\approx 0.333333333+0.19641855i$$

$$x=-\frac{5\sqrt{2}i}{36}+\frac{1}{3}\approx 0.333333333-0.19641855i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9x by 2-3x.

$$18x-27x^{2}=\frac{1}{2}+\frac{3}{8}+\frac{4}{6}+\frac{5}{2}$$

Least common multiple of 2 and 8 is 8. Convert \frac{1}{2} and \frac{3}{8} to fractions with denominator 8.

$$18x-27x^{2}=\frac{4}{8}+\frac{3}{8}+\frac{4}{6}+\frac{5}{2}$$

Since \frac{4}{8} and \frac{3}{8} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{4+3}{8}+\frac{4}{6}+\frac{5}{2}$$

Add 4 and 3 to get 7.

$$18x-27x^{2}=\frac{7}{8}+\frac{4}{6}+\frac{5}{2}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$18x-27x^{2}=\frac{7}{8}+\frac{2}{3}+\frac{5}{2}$$

Least common multiple of 8 and 3 is 24. Convert \frac{7}{8} and \frac{2}{3} to fractions with denominator 24.

$$18x-27x^{2}=\frac{21}{24}+\frac{16}{24}+\frac{5}{2}$$

Since \frac{21}{24} and \frac{16}{24} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{21+16}{24}+\frac{5}{2}$$

Add 21 and 16 to get 37.

$$18x-27x^{2}=\frac{37}{24}+\frac{5}{2}$$

Least common multiple of 24 and 2 is 24. Convert \frac{37}{24} and \frac{5}{2} to fractions with denominator 24.

$$18x-27x^{2}=\frac{37}{24}+\frac{60}{24}$$

Since \frac{37}{24} and \frac{60}{24} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{37+60}{24}$$

Add 37 and 60 to get 97.

$$18x-27x^{2}=\frac{97}{24}$$

Subtract \frac{97}{24} from both sides.

$$18x-27x^{2}-\frac{97}{24}=0$$

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.

$$-27x^{2}+18x-\frac{97}{24}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -27 for a, 18 for b, and -\frac{97}{24} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-27\right)\left(-\frac{97}{24}\right)}}{2\left(-27\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-27\right)\left(-\frac{97}{24}\right)}}{2\left(-27\right)}$$

Multiply -4 times -27.

$$x=\frac{-18±\sqrt{324+108\left(-\frac{97}{24}\right)}}{2\left(-27\right)}$$

Multiply 108 times -\frac{97}{24}.

$$x=\frac{-18±\sqrt{324-\frac{873}{2}}}{2\left(-27\right)}$$

Add 324 to -\frac{873}{2}.

$$x=\frac{-18±\sqrt{-\frac{225}{2}}}{2\left(-27\right)}$$

Take the square root of -\frac{225}{2}.

$$x=\frac{-18±\frac{15\sqrt{2}i}{2}}{2\left(-27\right)}$$

Multiply 2 times -27.

$$x=\frac{-18±\frac{15\sqrt{2}i}{2}}{-54}$$

Now solve the equation x=\frac{-18±\frac{15\sqrt{2}i}{2}}{-54} when ± is plus. Add -18 to \frac{15i\sqrt{2}}{2}.

$$x=\frac{\frac{15\sqrt{2}i}{2}-18}{-54}$$

Divide -18+\frac{15i\sqrt{2}}{2} by -54.

$$x=-\frac{5\sqrt{2}i}{36}+\frac{1}{3}$$

Now solve the equation x=\frac{-18±\frac{15\sqrt{2}i}{2}}{-54} when ± is minus. Subtract \frac{15i\sqrt{2}}{2} from -18.

$$x=\frac{-\frac{15\sqrt{2}i}{2}-18}{-54}$$

Divide -18-\frac{15i\sqrt{2}}{2} by -54.

$$x=\frac{5\sqrt{2}i}{36}+\frac{1}{3}$$

The equation is now solved.

$$x=-\frac{5\sqrt{2}i}{36}+\frac{1}{3} x=\frac{5\sqrt{2}i}{36}+\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 9x by 2-3x.

$$18x-27x^{2}=\frac{1}{2}+\frac{3}{8}+\frac{4}{6}+\frac{5}{2}$$

Least common multiple of 2 and 8 is 8. Convert \frac{1}{2} and \frac{3}{8} to fractions with denominator 8.

$$18x-27x^{2}=\frac{4}{8}+\frac{3}{8}+\frac{4}{6}+\frac{5}{2}$$

Since \frac{4}{8} and \frac{3}{8} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{4+3}{8}+\frac{4}{6}+\frac{5}{2}$$

Add 4 and 3 to get 7.

$$18x-27x^{2}=\frac{7}{8}+\frac{4}{6}+\frac{5}{2}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$18x-27x^{2}=\frac{7}{8}+\frac{2}{3}+\frac{5}{2}$$

Least common multiple of 8 and 3 is 24. Convert \frac{7}{8} and \frac{2}{3} to fractions with denominator 24.

$$18x-27x^{2}=\frac{21}{24}+\frac{16}{24}+\frac{5}{2}$$

Since \frac{21}{24} and \frac{16}{24} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{21+16}{24}+\frac{5}{2}$$

Add 21 and 16 to get 37.

$$18x-27x^{2}=\frac{37}{24}+\frac{5}{2}$$

Least common multiple of 24 and 2 is 24. Convert \frac{37}{24} and \frac{5}{2} to fractions with denominator 24.

$$18x-27x^{2}=\frac{37}{24}+\frac{60}{24}$$

Since \frac{37}{24} and \frac{60}{24} have the same denominator, add them by adding their numerators.

$$18x-27x^{2}=\frac{37+60}{24}$$

Add 37 and 60 to get 97.

$$18x-27x^{2}=\frac{97}{24}$$

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.

$$-27x^{2}+18x=\frac{97}{24}$$

Divide both sides by -27.

$$\frac{-27x^{2}+18x}{-27}=\frac{\frac{97}{24}}{-27}$$

Dividing by -27 undoes the multiplication by -27.

$$x^{2}+\frac{18}{-27}x=\frac{\frac{97}{24}}{-27}$$

Reduce the fraction \frac{18}{-27} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{2}{3}x=\frac{\frac{97}{24}}{-27}$$

Divide \frac{97}{24} by -27.

$$x^{2}-\frac{2}{3}x=-\frac{97}{648}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{97}{648}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{97}{648}+\frac{1}{9}$$

Add -\frac{97}{648} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{25}{648}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{25}{648}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{25}{648}}$$

Simplify.

$$x-\frac{1}{3}=\frac{5\sqrt{2}i}{36} x-\frac{1}{3}=-\frac{5\sqrt{2}i}{36}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{5\sqrt{2}i}{36}+\frac{1}{3} x=-\frac{5\sqrt{2}i}{36}+\frac{1}{3}$$