Solve 3(1-x)+4(1+2x)(1-x)=7

$$3(1-x)+4(1+2x)(1-x)=7$$

Solve for x

$$x=\frac{1}{8}=0.125$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 1-x.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

Use the distributive property to multiply 4 by 1+2x.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

Use the distributive property to multiply 4+8x by 1-x and combine like terms.

$$3-3x+4+4x-8x^{2}=7$$

Add 3 and 4 to get 7.

$$7-3x+4x-8x^{2}=7$$

Combine -3x and 4x to get x.

$$7+x-8x^{2}=7$$

Subtract 7 from both sides.

$$7+x-8x^{2}-7=0$$

Subtract 7 from 7 to get 0.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-1±1}{-16}$$

Now solve the equation x=\frac{-1±1}{-16} when ± is plus. Add -1 to 1.

$$x=\frac{0}{-16}$$

Divide 0 by -16.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-16} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-16}$$

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

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=0 x=\frac{1}{8}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by 1-x.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

Use the distributive property to multiply 4 by 1+2x.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

Use the distributive property to multiply 4+8x by 1-x and combine like terms.

$$3-3x+4+4x-8x^{2}=7$$

Add 3 and 4 to get 7.

$$7-3x+4x-8x^{2}=7$$

Combine -3x and 4x to get x.

$$7+x-8x^{2}=7$$

Subtract 7 from both sides.

$$x-8x^{2}=7-7$$

Subtract 7 from 7 to get 0.

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

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.

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

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{1}{-8}x=\frac{0}{-8}$$

Divide 1 by -8.

$$x^{2}-\frac{1}{8}x=\frac{0}{-8}$$

Divide 0 by -8.

$$x^{2}-\frac{1}{8}x=0$$

Divide -\frac{1}{8}, the coefficient of the x term, by 2 to get -\frac{1}{16}. Then add the square of -\frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\left(-\frac{1}{16}\right)^{2}$$

Square -\frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

Factor x^{2}-\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{1}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Simplify.

$$x-\frac{1}{16}=\frac{1}{16} x-\frac{1}{16}=-\frac{1}{16}$$

Add \frac{1}{16} to both sides of the equation.

$$x=\frac{1}{8} x=0$$