Solve x-0.5-1/18x^2=(9-x)(1-1/9x)*1/2

$$x-0.5- \frac{ 1 }{ 18 } { x }^{ 2 } =(9-x)(1- \frac{ 1 }{ 9 } x) \times \frac{ 1 }{ 2 }$$

Solve for x

$$x=3$$

$$x=15$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9-x by 1-\frac{1}{9}x and combine like terms.

$$x-0.5-\frac{1}{18}x^{2}=\left(9-2x+\frac{1}{9}x^{2}\right)\times \frac{1}{2}$$

Use the distributive property to multiply 9-2x+\frac{1}{9}x^{2} by \frac{1}{2}.

$$x-0.5-\frac{1}{18}x^{2}=\frac{9}{2}-x+\frac{1}{18}x^{2}$$

Subtract \frac{9}{2} from both sides.

$$x-0.5-\frac{1}{18}x^{2}-\frac{9}{2}=-x+\frac{1}{18}x^{2}$$

Subtract \frac{9}{2} from -0.5 to get -5.

$$x-5-\frac{1}{18}x^{2}=-x+\frac{1}{18}x^{2}$$

Add x to both sides.

$$x-5-\frac{1}{18}x^{2}+x=\frac{1}{18}x^{2}$$

Combine x and x to get 2x.

$$2x-5-\frac{1}{18}x^{2}=\frac{1}{18}x^{2}$$

Subtract \frac{1}{18}x^{2} from both sides.

$$2x-5-\frac{1}{18}x^{2}-\frac{1}{18}x^{2}=0$$

Combine -\frac{1}{18}x^{2} and -\frac{1}{18}x^{2} to get -\frac{1}{9}x^{2}.

$$2x-5-\frac{1}{9}x^{2}=0$$

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.

$$-\frac{1}{9}x^{2}+2x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{9} for a, 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-\frac{1}{9}\right)\left(-5\right)}}{2\left(-\frac{1}{9}\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{1}{9}\right)\left(-5\right)}}{2\left(-\frac{1}{9}\right)}$$

Multiply -4 times -\frac{1}{9}.

$$x=\frac{-2±\sqrt{4+\frac{4}{9}\left(-5\right)}}{2\left(-\frac{1}{9}\right)}$$

Multiply \frac{4}{9} times -5.

$$x=\frac{-2±\sqrt{4-\frac{20}{9}}}{2\left(-\frac{1}{9}\right)}$$

Add 4 to -\frac{20}{9}.

$$x=\frac{-2±\sqrt{\frac{16}{9}}}{2\left(-\frac{1}{9}\right)}$$

Take the square root of \frac{16}{9}.

$$x=\frac{-2±\frac{4}{3}}{2\left(-\frac{1}{9}\right)}$$

Multiply 2 times -\frac{1}{9}.

$$x=\frac{-2±\frac{4}{3}}{-\frac{2}{9}}$$

Now solve the equation x=\frac{-2±\frac{4}{3}}{-\frac{2}{9}} when ± is plus. Add -2 to \frac{4}{3}.

$$x=-\frac{\frac{2}{3}}{-\frac{2}{9}}$$

Divide -\frac{2}{3} by -\frac{2}{9} by multiplying -\frac{2}{3} by the reciprocal of -\frac{2}{9}.

$$x=3$$

Now solve the equation x=\frac{-2±\frac{4}{3}}{-\frac{2}{9}} when ± is minus. Subtract \frac{4}{3} from -2.

$$x=-\frac{\frac{10}{3}}{-\frac{2}{9}}$$

Divide -\frac{10}{3} by -\frac{2}{9} by multiplying -\frac{10}{3} by the reciprocal of -\frac{2}{9}.

$$x=15$$

The equation is now solved.

$$x=3 x=15$$

Steps for Completing the Square

Use the distributive property to multiply 9-x by 1-\frac{1}{9}x and combine like terms.

$$x-0.5-\frac{1}{18}x^{2}=\left(9-2x+\frac{1}{9}x^{2}\right)\times \frac{1}{2}$$

Use the distributive property to multiply 9-2x+\frac{1}{9}x^{2} by \frac{1}{2}.

$$x-0.5-\frac{1}{18}x^{2}=\frac{9}{2}-x+\frac{1}{18}x^{2}$$

Add x to both sides.

$$x-0.5-\frac{1}{18}x^{2}+x=\frac{9}{2}+\frac{1}{18}x^{2}$$

Combine x and x to get 2x.

$$2x-0.5-\frac{1}{18}x^{2}=\frac{9}{2}+\frac{1}{18}x^{2}$$

Subtract \frac{1}{18}x^{2} from both sides.

$$2x-0.5-\frac{1}{18}x^{2}-\frac{1}{18}x^{2}=\frac{9}{2}$$

Combine -\frac{1}{18}x^{2} and -\frac{1}{18}x^{2} to get -\frac{1}{9}x^{2}.

$$2x-0.5-\frac{1}{9}x^{2}=\frac{9}{2}$$

Add 0.5 to both sides.

$$2x-\frac{1}{9}x^{2}=\frac{9}{2}+0.5$$

Add \frac{9}{2} and 0.5 to get 5.

$$2x-\frac{1}{9}x^{2}=5$$

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.

$$-\frac{1}{9}x^{2}+2x=5$$

Multiply both sides by -9.

$$\frac{-\frac{1}{9}x^{2}+2x}{-\frac{1}{9}}=\frac{5}{-\frac{1}{9}}$$

Dividing by -\frac{1}{9} undoes the multiplication by -\frac{1}{9}.

$$x^{2}+\frac{2}{-\frac{1}{9}}x=\frac{5}{-\frac{1}{9}}$$

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

$$x^{2}-18x=\frac{5}{-\frac{1}{9}}$$

Divide 5 by -\frac{1}{9} by multiplying 5 by the reciprocal of -\frac{1}{9}.

$$x^{2}-18x=-45$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=-45+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=-45+81$$

Add -45 to 81.

$$x^{2}-18x+81=36$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-9=6 x-9=-6$$

Add 9 to both sides of the equation.

$$x=15 x=3$$