Solve 1-(2-a)^2=(1-a/2)^2

$$1 - ( 2 - a ) ^ { 2 } = ( 1 - \frac { a } { 2 } ) ^ { 2 }$$

Solve for a

$$a=-\frac{2\sqrt{5}}{5}+2\approx 1.105572809$$

$$a=\frac{2\sqrt{5}}{5}+2\approx 2.894427191$$

Steps Using the Quadratic Formula

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

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

To find the opposite of 4-4a+a^{2}, find the opposite of each term.

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

Subtract 4 from 1 to get -3.

$$-3+4a-a^{2}=\left(1-\frac{a}{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{a}{2}\right)^{2}.

$$-3+4a-a^{2}=1+2\left(-\frac{a}{2}\right)+\left(-\frac{a}{2}\right)^{2}$$

Express 2\left(-\frac{a}{2}\right) as a single fraction.

$$-3+4a-a^{2}=1+\frac{-2a}{2}+\left(-\frac{a}{2}\right)^{2}$$

Cancel out 2 and 2.

$$-3+4a-a^{2}=1-a+\left(-\frac{a}{2}\right)^{2}$$

Calculate -\frac{a}{2} to the power of 2 and get \left(\frac{a}{2}\right)^{2}.

$$-3+4a-a^{2}=1-a+\left(\frac{a}{2}\right)^{2}$$

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-3+4a-a^{2}=1-a+\frac{a^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1-a times \frac{2^{2}}{2^{2}}.

$$-3+4a-a^{2}=\frac{\left(1-a\right)\times 2^{2}}{2^{2}}+\frac{a^{2}}{2^{2}}$$

Since \frac{\left(1-a\right)\times 2^{2}}{2^{2}} and \frac{a^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$-3+4a-a^{2}=\frac{\left(1-a\right)\times 2^{2}+a^{2}}{2^{2}}$$

Do the multiplications in \left(1-a\right)\times 2^{2}+a^{2}.

$$-3+4a-a^{2}=\frac{4-4a+a^{2}}{2^{2}}$$

Subtract \frac{4-4a+a^{2}}{2^{2}} from both sides.

$$-3+4a-a^{2}-\frac{4-4a+a^{2}}{2^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$-3+4a-a^{2}-\frac{4-4a+a^{2}}{4}=0$$

Divide each term of 4-4a+a^{2} by 4 to get 1-a+\frac{1}{4}a^{2}.

$$-3+4a-a^{2}-\left(1-a+\frac{1}{4}a^{2}\right)=0$$

To find the opposite of 1-a+\frac{1}{4}a^{2}, find the opposite of each term.

$$-3+4a-a^{2}-1+a-\frac{1}{4}a^{2}=0$$

Subtract 1 from -3 to get -4.

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

Combine 4a and a to get 5a.

$$-4+5a-a^{2}-\frac{1}{4}a^{2}=0$$

Combine -a^{2} and -\frac{1}{4}a^{2} to get -\frac{5}{4}a^{2}.

$$-4+5a-\frac{5}{4}a^{2}=0$$

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.

$$-\frac{5}{4}a^{2}+5a-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{4} for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-5±\sqrt{5^{2}-4\left(-\frac{5}{4}\right)\left(-4\right)}}{2\left(-\frac{5}{4}\right)}$$

Square 5.

$$a=\frac{-5±\sqrt{25-4\left(-\frac{5}{4}\right)\left(-4\right)}}{2\left(-\frac{5}{4}\right)}$$

Multiply -4 times -\frac{5}{4}.

$$a=\frac{-5±\sqrt{25+5\left(-4\right)}}{2\left(-\frac{5}{4}\right)}$$

Multiply 5 times -4.

$$a=\frac{-5±\sqrt{25-20}}{2\left(-\frac{5}{4}\right)}$$

Add 25 to -20.

$$a=\frac{-5±\sqrt{5}}{2\left(-\frac{5}{4}\right)}$$

Multiply 2 times -\frac{5}{4}.

$$a=\frac{-5±\sqrt{5}}{-\frac{5}{2}}$$

Now solve the equation a=\frac{-5±\sqrt{5}}{-\frac{5}{2}} when ± is plus. Add -5 to \sqrt{5}.

$$a=\frac{\sqrt{5}-5}{-\frac{5}{2}}$$

Divide -5+\sqrt{5} by -\frac{5}{2} by multiplying -5+\sqrt{5} by the reciprocal of -\frac{5}{2}.

$$a=-\frac{2\sqrt{5}}{5}+2$$

Now solve the equation a=\frac{-5±\sqrt{5}}{-\frac{5}{2}} when ± is minus. Subtract \sqrt{5} from -5.

$$a=\frac{-\sqrt{5}-5}{-\frac{5}{2}}$$

Divide -5-\sqrt{5} by -\frac{5}{2} by multiplying -5-\sqrt{5} by the reciprocal of -\frac{5}{2}.

$$a=\frac{2\sqrt{5}}{5}+2$$

The equation is now solved.

$$a=-\frac{2\sqrt{5}}{5}+2 a=\frac{2\sqrt{5}}{5}+2$$

Steps for Completing the Square

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

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

To find the opposite of 4-4a+a^{2}, find the opposite of each term.

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

Subtract 4 from 1 to get -3.

$$-3+4a-a^{2}=\left(1-\frac{a}{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{a}{2}\right)^{2}.

$$-3+4a-a^{2}=1+2\left(-\frac{a}{2}\right)+\left(-\frac{a}{2}\right)^{2}$$

Express 2\left(-\frac{a}{2}\right) as a single fraction.

$$-3+4a-a^{2}=1+\frac{-2a}{2}+\left(-\frac{a}{2}\right)^{2}$$

Cancel out 2 and 2.

$$-3+4a-a^{2}=1-a+\left(-\frac{a}{2}\right)^{2}$$

Calculate -\frac{a}{2} to the power of 2 and get \left(\frac{a}{2}\right)^{2}.

$$-3+4a-a^{2}=1-a+\left(\frac{a}{2}\right)^{2}$$

To raise \frac{a}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-3+4a-a^{2}=1-a+\frac{a^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1-a times \frac{2^{2}}{2^{2}}.

$$-3+4a-a^{2}=\frac{\left(1-a\right)\times 2^{2}}{2^{2}}+\frac{a^{2}}{2^{2}}$$

Since \frac{\left(1-a\right)\times 2^{2}}{2^{2}} and \frac{a^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$-3+4a-a^{2}=\frac{\left(1-a\right)\times 2^{2}+a^{2}}{2^{2}}$$

Do the multiplications in \left(1-a\right)\times 2^{2}+a^{2}.

$$-3+4a-a^{2}=\frac{4-4a+a^{2}}{2^{2}}$$

Subtract \frac{4-4a+a^{2}}{2^{2}} from both sides.

$$-3+4a-a^{2}-\frac{4-4a+a^{2}}{2^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$-3+4a-a^{2}-\frac{4-4a+a^{2}}{4}=0$$

Divide each term of 4-4a+a^{2} by 4 to get 1-a+\frac{1}{4}a^{2}.

$$-3+4a-a^{2}-\left(1-a+\frac{1}{4}a^{2}\right)=0$$

To find the opposite of 1-a+\frac{1}{4}a^{2}, find the opposite of each term.

$$-3+4a-a^{2}-1+a-\frac{1}{4}a^{2}=0$$

Subtract 1 from -3 to get -4.

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

Combine 4a and a to get 5a.

$$-4+5a-a^{2}-\frac{1}{4}a^{2}=0$$

Combine -a^{2} and -\frac{1}{4}a^{2} to get -\frac{5}{4}a^{2}.

$$-4+5a-\frac{5}{4}a^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$5a-\frac{5}{4}a^{2}=4$$

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.

$$-\frac{5}{4}a^{2}+5a=4$$

Divide both sides of the equation by -\frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{4}a^{2}+5a}{-\frac{5}{4}}=\frac{4}{-\frac{5}{4}}$$

Dividing by -\frac{5}{4} undoes the multiplication by -\frac{5}{4}.

$$a^{2}+\frac{5}{-\frac{5}{4}}a=\frac{4}{-\frac{5}{4}}$$

Divide 5 by -\frac{5}{4} by multiplying 5 by the reciprocal of -\frac{5}{4}.

$$a^{2}-4a=\frac{4}{-\frac{5}{4}}$$

Divide 4 by -\frac{5}{4} by multiplying 4 by the reciprocal of -\frac{5}{4}.

$$a^{2}-4a=-\frac{16}{5}$$

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

$$a^{2}-4a+\left(-2\right)^{2}=-\frac{16}{5}+\left(-2\right)^{2}$$

Square -2.

$$a^{2}-4a+4=-\frac{16}{5}+4$$

Add -\frac{16}{5} to 4.

$$a^{2}-4a+4=\frac{4}{5}$$

Factor a^{2}-4a+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}.

$$\left(a-2\right)^{2}=\frac{4}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{\frac{4}{5}}$$

Simplify.

$$a-2=\frac{2\sqrt{5}}{5} a-2=-\frac{2\sqrt{5}}{5}$$

Add 2 to both sides of the equation.

$$a=\frac{2\sqrt{5}}{5}+2 a=-\frac{2\sqrt{5}}{5}+2$$