Solve left(a+2right)^2=left(4-aright)^2+a^2

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

Solve for a

$$a=2\sqrt{6}+6\approx 10.898979486$$

$$a=6-2\sqrt{6}\approx 1.101020514$$

Steps Using the Quadratic Formula

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$a^{2}+4a+4=16-8a+a^{2}+a^{2}$$

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

$$a^{2}+4a+4=16-8a+2a^{2}$$

Subtract 16 from both sides.

$$a^{2}+4a+4-16=-8a+2a^{2}$$

Subtract 16 from 4 to get -12.

$$a^{2}+4a-12=-8a+2a^{2}$$

Add 8a to both sides.

$$a^{2}+4a-12+8a=2a^{2}$$

Combine 4a and 8a to get 12a.

$$a^{2}+12a-12=2a^{2}$$

Subtract 2a^{2} from both sides.

$$a^{2}+12a-12-2a^{2}=0$$

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

$$-a^{2}+12a-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 12 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-12±\sqrt{12^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$a=\frac{-12±\sqrt{144+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$a=\frac{-12±\sqrt{144-48}}{2\left(-1\right)}$$

Add 144 to -48.

$$a=\frac{-12±\sqrt{96}}{2\left(-1\right)}$$

Take the square root of 96.

$$a=\frac{-12±4\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$a=\frac{-12±4\sqrt{6}}{-2}$$

Now solve the equation a=\frac{-12±4\sqrt{6}}{-2} when ± is plus. Add -12 to 4\sqrt{6}.

$$a=\frac{4\sqrt{6}-12}{-2}$$

Divide -12+4\sqrt{6} by -2.

$$a=6-2\sqrt{6}$$

Now solve the equation a=\frac{-12±4\sqrt{6}}{-2} when ± is minus. Subtract 4\sqrt{6} from -12.

$$a=\frac{-4\sqrt{6}-12}{-2}$$

Divide -12-4\sqrt{6} by -2.

$$a=2\sqrt{6}+6$$

The equation is now solved.

$$a=6-2\sqrt{6} a=2\sqrt{6}+6$$

Steps for Completing the Square

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-a\right)^{2}.

$$a^{2}+4a+4=16-8a+a^{2}+a^{2}$$

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

$$a^{2}+4a+4=16-8a+2a^{2}$$

Add 8a to both sides.

$$a^{2}+4a+4+8a=16+2a^{2}$$

Combine 4a and 8a to get 12a.

$$a^{2}+12a+4=16+2a^{2}$$

Subtract 2a^{2} from both sides.

$$a^{2}+12a+4-2a^{2}=16$$

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

$$-a^{2}+12a+4=16$$

Subtract 4 from both sides.

$$-a^{2}+12a=16-4$$

Subtract 4 from 16 to get 12.

$$-a^{2}+12a=12$$

Divide both sides by -1.

$$\frac{-a^{2}+12a}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a^{2}+\frac{12}{-1}a=\frac{12}{-1}$$

Divide 12 by -1.

$$a^{2}-12a=\frac{12}{-1}$$

Divide 12 by -1.

$$a^{2}-12a=-12$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-12a+\left(-6\right)^{2}=-12+\left(-6\right)^{2}$$

Square -6.

$$a^{2}-12a+36=-12+36$$

Add -12 to 36.

$$a^{2}-12a+36=24$$

Factor a^{2}-12a+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-6\right)^{2}=24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-6\right)^{2}}=\sqrt{24}$$

Simplify.

$$a-6=2\sqrt{6} a-6=-2\sqrt{6}$$

Add 6 to both sides of the equation.

$$a=2\sqrt{6}+6 a=6-2\sqrt{6}$$