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

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

Solve for a

$$a=\frac{7+4\sqrt{2}i}{9}\approx 0.777777778+0.628539361i$$

$$a=\frac{-4\sqrt{2}i+7}{9}\approx 0.777777778-0.628539361i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

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

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

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

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

Use the distributive property to multiply 2 by 4-4a+a^{2}.

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

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

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

Use the distributive property to multiply 2 by 4a^{2}-4a+1.

$$1-2a+a^{2}=8-8a+2a^{2}+8a^{2}-8a+2$$

Combine 2a^{2} and 8a^{2} to get 10a^{2}.

$$1-2a+a^{2}=8-8a+10a^{2}-8a+2$$

Combine -8a and -8a to get -16a.

$$1-2a+a^{2}=8-16a+10a^{2}+2$$

Add 8 and 2 to get 10.

$$1-2a+a^{2}=10-16a+10a^{2}$$

Subtract 10 from both sides.

$$1-2a+a^{2}-10=-16a+10a^{2}$$

Subtract 10 from 1 to get -9.

$$-9-2a+a^{2}=-16a+10a^{2}$$

Add 16a to both sides.

$$-9-2a+a^{2}+16a=10a^{2}$$

Combine -2a and 16a to get 14a.

$$-9+14a+a^{2}=10a^{2}$$

Subtract 10a^{2} from both sides.

$$-9+14a+a^{2}-10a^{2}=0$$

Combine a^{2} and -10a^{2} to get -9a^{2}.

$$-9+14a-9a^{2}=0$$

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.

$$-9a^{2}+14a-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 14 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-14±\sqrt{14^{2}-4\left(-9\right)\left(-9\right)}}{2\left(-9\right)}$$

Square 14.

$$a=\frac{-14±\sqrt{196-4\left(-9\right)\left(-9\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$a=\frac{-14±\sqrt{196+36\left(-9\right)}}{2\left(-9\right)}$$

Multiply 36 times -9.

$$a=\frac{-14±\sqrt{196-324}}{2\left(-9\right)}$$

Add 196 to -324.

$$a=\frac{-14±\sqrt{-128}}{2\left(-9\right)}$$

Take the square root of -128.

$$a=\frac{-14±8\sqrt{2}i}{2\left(-9\right)}$$

Multiply 2 times -9.

$$a=\frac{-14±8\sqrt{2}i}{-18}$$

Now solve the equation a=\frac{-14±8\sqrt{2}i}{-18} when ± is plus. Add -14 to 8i\sqrt{2}.

$$a=\frac{-14+8\sqrt{2}i}{-18}$$

Divide -14+8i\sqrt{2} by -18.

$$a=\frac{-4\sqrt{2}i+7}{9}$$

Now solve the equation a=\frac{-14±8\sqrt{2}i}{-18} when ± is minus. Subtract 8i\sqrt{2} from -14.

$$a=\frac{-8\sqrt{2}i-14}{-18}$$

Divide -14-8i\sqrt{2} by -18.

$$a=\frac{7+4\sqrt{2}i}{9}$$

The equation is now solved.

$$a=\frac{-4\sqrt{2}i+7}{9} a=\frac{7+4\sqrt{2}i}{9}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

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

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

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

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

Use the distributive property to multiply 2 by 4-4a+a^{2}.

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

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

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

Use the distributive property to multiply 2 by 4a^{2}-4a+1.

$$1-2a+a^{2}=8-8a+2a^{2}+8a^{2}-8a+2$$

Combine 2a^{2} and 8a^{2} to get 10a^{2}.

$$1-2a+a^{2}=8-8a+10a^{2}-8a+2$$

Combine -8a and -8a to get -16a.

$$1-2a+a^{2}=8-16a+10a^{2}+2$$

Add 8 and 2 to get 10.

$$1-2a+a^{2}=10-16a+10a^{2}$$

Add 16a to both sides.

$$1-2a+a^{2}+16a=10+10a^{2}$$

Combine -2a and 16a to get 14a.

$$1+14a+a^{2}=10+10a^{2}$$

Subtract 10a^{2} from both sides.

$$1+14a+a^{2}-10a^{2}=10$$

Combine a^{2} and -10a^{2} to get -9a^{2}.

$$1+14a-9a^{2}=10$$

Subtract 1 from both sides.

$$14a-9a^{2}=10-1$$

Subtract 1 from 10 to get 9.

$$14a-9a^{2}=9$$

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.

$$-9a^{2}+14a=9$$

Divide both sides by -9.

$$\frac{-9a^{2}+14a}{-9}=\frac{9}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$a^{2}+\frac{14}{-9}a=\frac{9}{-9}$$

Divide 14 by -9.

$$a^{2}-\frac{14}{9}a=\frac{9}{-9}$$

Divide 9 by -9.

$$a^{2}-\frac{14}{9}a=-1$$

Divide -\frac{14}{9}, the coefficient of the x term, by 2 to get -\frac{7}{9}. Then add the square of -\frac{7}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{14}{9}a+\left(-\frac{7}{9}\right)^{2}=-1+\left(-\frac{7}{9}\right)^{2}$$

Square -\frac{7}{9} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{14}{9}a+\frac{49}{81}=-1+\frac{49}{81}$$

Add -1 to \frac{49}{81}.

$$a^{2}-\frac{14}{9}a+\frac{49}{81}=-\frac{32}{81}$$

Factor a^{2}-\frac{14}{9}a+\frac{49}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{7}{9}\right)^{2}=-\frac{32}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{7}{9}\right)^{2}}=\sqrt{-\frac{32}{81}}$$

Simplify.

$$a-\frac{7}{9}=\frac{4\sqrt{2}i}{9} a-\frac{7}{9}=-\frac{4\sqrt{2}i}{9}$$

Add \frac{7}{9} to both sides of the equation.

$$a=\frac{7+4\sqrt{2}i}{9} a=\frac{-4\sqrt{2}i+7}{9}$$