Solve {l}{x=3(3x(1-x))(1-3x(1-x)}{=9x(1-x)(1-(3x-3x^2)))}

$$\left. \begin{array} { l } { x = 3 ( 3 x ( 1 - x ) ) ( 1 - 3 x ( 1 - x ) } \\ { = 9 x ( 1 - x ) ( 1 - ( 3 x - 3 x ^ { 2 } ) ) ) } \end{array} \right.$$

Solve for x

$$x=\frac{8}{9}\approx 0.888888889$$

$$x=0$$

Steps Using Factoring

Multiply 3 and 3 to get 9.

$$x=9x\left(1-x\right)$$

Use the distributive property to multiply 9x by 1-x.

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

Subtract 9x from both sides.

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

Combine x and -9x to get -8x.

$$-8x=-9x^{2}$$

Add 9x^{2} to both sides.

$$-8x+9x^{2}=0$$

Factor out x.

$$x\left(-8+9x\right)=0$$

To find equation solutions, solve x=0 and -8+9x=0.

$$x=0 x=\frac{8}{9}$$

Steps Using the Quadratic Formula

Multiply 3 and 3 to get 9.

$$x=9x\left(1-x\right)$$

Use the distributive property to multiply 9x by 1-x.

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

Subtract 9x from both sides.

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

Combine x and -9x to get -8x.

$$-8x=-9x^{2}$$

Add 9x^{2} to both sides.

$$-8x+9x^{2}=0$$

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.

$$9x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 9}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 9}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{8±8}{18}$$

Now solve the equation x=\frac{8±8}{18} when ± is plus. Add 8 to 8.

$$x=\frac{16}{18}$$

Reduce the fraction \frac{16}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{9}$$

Now solve the equation x=\frac{8±8}{18} when ± is minus. Subtract 8 from 8.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

$$x=\frac{8}{9} x=0$$

Steps for Completing the Square

Multiply 3 and 3 to get 9.

$$x=9x\left(1-x\right)$$

Use the distributive property to multiply 9x by 1-x.

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

Subtract 9x from both sides.

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

Combine x and -9x to get -8x.

$$-8x=-9x^{2}$$

Add 9x^{2} to both sides.

$$-8x+9x^{2}=0$$

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.

$$9x^{2}-8x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-8x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{8}{9}x=\frac{0}{9}$$

Divide 0 by 9.

$$x^{2}-\frac{8}{9}x=0$$

Divide -\frac{8}{9}, the coefficient of the x term, by 2 to get -\frac{4}{9}. Then add the square of -\frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{9}x+\left(-\frac{4}{9}\right)^{2}=\left(-\frac{4}{9}\right)^{2}$$

Square -\frac{4}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{9}x+\frac{16}{81}=\frac{16}{81}$$

Factor x^{2}-\frac{8}{9}x+\frac{16}{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}.

$$\left(x-\frac{4}{9}\right)^{2}=\frac{16}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{9}\right)^{2}}=\sqrt{\frac{16}{81}}$$

Simplify.

$$x-\frac{4}{9}=\frac{4}{9} x-\frac{4}{9}=-\frac{4}{9}$$

Add \frac{4}{9} to both sides of the equation.

$$x=\frac{8}{9} x=0$$