Solve 2x(1-x)=1/4

$$2 x ( 1 - x ) = \frac { 1 } { 4 }$$

Solve for x

$$x=\frac{\sqrt{2}}{4}+\frac{1}{2}\approx 0.853553391$$

$$x=-\frac{\sqrt{2}}{4}+\frac{1}{2}\approx 0.146446609$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by 1-x.

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

Subtract \frac{1}{4} from both sides.

$$2x-2x^{2}-\frac{1}{4}=0$$

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.

$$-2x^{2}+2x-\frac{1}{4}=0$$

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

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

Square 2.

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

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8\left(-\frac{1}{4}\right)}}{2\left(-2\right)}$$

Multiply 8 times -\frac{1}{4}.

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

Add 4 to -2.

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

Multiply 2 times -2.

$$x=\frac{-2±\sqrt{2}}{-4}$$

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

$$x=\frac{\sqrt{2}-2}{-4}$$

Divide -2+\sqrt{2} by -4.

$$x=-\frac{\sqrt{2}}{4}+\frac{1}{2}$$

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

$$x=\frac{-\sqrt{2}-2}{-4}$$

Divide -2-\sqrt{2} by -4.

$$x=\frac{\sqrt{2}}{4}+\frac{1}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{2}}{4}+\frac{1}{2} x=\frac{\sqrt{2}}{4}+\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 2x by 1-x.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

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

Divide \frac{1}{4} by -2.

$$x^{2}-x=-\frac{1}{8}$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{8}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{1}{8}+\frac{1}{4}$$

Add -\frac{1}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{1}{8}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{1}{8}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{8}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{2}}{4} x-\frac{1}{2}=-\frac{\sqrt{2}}{4}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{2}}{4}+\frac{1}{2} x=-\frac{\sqrt{2}}{4}+\frac{1}{2}$$