Solve (1-x)(1-2x)=432/600

$$( 1 - x ) ( 1 - 2 x ) = \frac { 432 } { 600 }$$

Solve for x

$$x=\frac{1}{10}=0.1$$

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1-x by 1-2x and combine like terms.

$$1-3x+2x^{2}=\frac{432}{600}$$

Reduce the fraction \frac{432}{600} to lowest terms by extracting and canceling out 24.

$$1-3x+2x^{2}=\frac{18}{25}$$

Subtract \frac{18}{25} from both sides.

$$1-3x+2x^{2}-\frac{18}{25}=0$$

Subtract \frac{18}{25} from 1 to get \frac{7}{25}.

$$\frac{7}{25}-3x+2x^{2}=0$$

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.

$$2x^{2}-3x+\frac{7}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and \frac{7}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times \frac{7}{25}}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times \frac{7}{25}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times \frac{7}{25}}}{2\times 2}$$

Multiply -8 times \frac{7}{25}.

$$x=\frac{-\left(-3\right)±\sqrt{9-\frac{56}{25}}}{2\times 2}$$

Add 9 to -\frac{56}{25}.

$$x=\frac{-\left(-3\right)±\sqrt{\frac{169}{25}}}{2\times 2}$$

Take the square root of \frac{169}{25}.

$$x=\frac{-\left(-3\right)±\frac{13}{5}}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±\frac{13}{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±\frac{13}{5}}{4}$$

Now solve the equation x=\frac{3±\frac{13}{5}}{4} when ± is plus. Add 3 to \frac{13}{5}.

$$x=\frac{\frac{28}{5}}{4}$$

Divide \frac{28}{5} by 4.

$$x=\frac{7}{5}$$

Now solve the equation x=\frac{3±\frac{13}{5}}{4} when ± is minus. Subtract \frac{13}{5} from 3.

$$x=\frac{\frac{2}{5}}{4}$$

Divide \frac{2}{5} by 4.

$$x=\frac{1}{10}$$

The equation is now solved.

$$x=\frac{7}{5} x=\frac{1}{10}$$

Steps for Completing the Square

Use the distributive property to multiply 1-x by 1-2x and combine like terms.

$$1-3x+2x^{2}=\frac{432}{600}$$

Reduce the fraction \frac{432}{600} to lowest terms by extracting and canceling out 24.

$$1-3x+2x^{2}=\frac{18}{25}$$

Subtract 1 from both sides.

$$-3x+2x^{2}=\frac{18}{25}-1$$

Subtract 1 from \frac{18}{25} to get -\frac{7}{25}.

$$-3x+2x^{2}=-\frac{7}{25}$$

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.

$$2x^{2}-3x=-\frac{7}{25}$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=-\frac{\frac{7}{25}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=-\frac{\frac{7}{25}}{2}$$

Divide -\frac{7}{25} by 2.

$$x^{2}-\frac{3}{2}x=-\frac{7}{50}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{7}{50}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{7}{50}+\frac{9}{16}$$

Add -\frac{7}{50} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{169}{400}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{169}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{169}{400}}$$

Simplify.

$$x-\frac{3}{4}=\frac{13}{20} x-\frac{3}{4}=-\frac{13}{20}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{7}{5} x=\frac{1}{10}$$