Solve (2-4a)(2+a)=(2-2a)^2

$$( 2 - 4 a ) ( 2 + a ) = ( 2 - 2 a ) ^ { 2 }$$

Solve for a

$$a=\frac{1}{4}=0.25$$

$$a=0$$

Steps Using Factoring

Use the distributive property to multiply 2-4a by 2+a and combine like terms.

$$4-6a-4a^{2}=\left(2-2a\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2a\right)^{2}.

$$4-6a-4a^{2}=4-8a+4a^{2}$$

Subtract 4 from both sides.

$$4-6a-4a^{2}-4=-8a+4a^{2}$$

Subtract 4 from 4 to get 0.

$$-6a-4a^{2}=-8a+4a^{2}$$

Add 8a to both sides.

$$-6a-4a^{2}+8a=4a^{2}$$

Combine -6a and 8a to get 2a.

$$2a-4a^{2}=4a^{2}$$

Subtract 4a^{2} from both sides.

$$2a-4a^{2}-4a^{2}=0$$

Combine -4a^{2} and -4a^{2} to get -8a^{2}.

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

Factor out a.

$$a\left(2-8a\right)=0$$

To find equation solutions, solve a=0 and 2-8a=0.

$$a=0 a=\frac{1}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2-4a by 2+a and combine like terms.

$$4-6a-4a^{2}=\left(2-2a\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2a\right)^{2}.

$$4-6a-4a^{2}=4-8a+4a^{2}$$

Subtract 4 from both sides.

$$4-6a-4a^{2}-4=-8a+4a^{2}$$

Subtract 4 from 4 to get 0.

$$-6a-4a^{2}=-8a+4a^{2}$$

Add 8a to both sides.

$$-6a-4a^{2}+8a=4a^{2}$$

Combine -6a and 8a to get 2a.

$$2a-4a^{2}=4a^{2}$$

Subtract 4a^{2} from both sides.

$$2a-4a^{2}-4a^{2}=0$$

Combine -4a^{2} and -4a^{2} to get -8a^{2}.

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

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.

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

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

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

Take the square root of 2^{2}.

$$a=\frac{-2±2}{2\left(-8\right)}$$

Multiply 2 times -8.

$$a=\frac{-2±2}{-16}$$

Now solve the equation a=\frac{-2±2}{-16} when ± is plus. Add -2 to 2.

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

Divide 0 by -16.

$$a=0$$

Now solve the equation a=\frac{-2±2}{-16} when ± is minus. Subtract 2 from -2.

$$a=-\frac{4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out 4.

$$a=\frac{1}{4}$$

The equation is now solved.

$$a=0 a=\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2-4a by 2+a and combine like terms.

$$4-6a-4a^{2}=\left(2-2a\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-2a\right)^{2}.

$$4-6a-4a^{2}=4-8a+4a^{2}$$

Add 8a to both sides.

$$4-6a-4a^{2}+8a=4+4a^{2}$$

Combine -6a and 8a to get 2a.

$$4+2a-4a^{2}=4+4a^{2}$$

Subtract 4a^{2} from both sides.

$$4+2a-4a^{2}-4a^{2}=4$$

Combine -4a^{2} and -4a^{2} to get -8a^{2}.

$$4+2a-8a^{2}=4$$

Subtract 4 from both sides.

$$2a-8a^{2}=4-4$$

Subtract 4 from 4 to get 0.

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

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.

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

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

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

Reduce the fraction \frac{2}{-8} to lowest terms by extracting and canceling out 2.

$$a^{2}-\frac{1}{4}a=\frac{0}{-8}$$

Divide 0 by -8.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{1}{4}a+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{1}{4}a+\frac{1}{64}=\frac{1}{64}$$

Factor a^{2}-\frac{1}{4}a+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$a-\frac{1}{8}=\frac{1}{8} a-\frac{1}{8}=-\frac{1}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$a=\frac{1}{4} a=0$$