Solve (4+2a)^2+(a+2)^2=16

$$( 4 + 2 a ) ^ { 2 } + ( a + 2 ) ^ { 2 } = 16$$

Solve for a

$$a=\frac{4\sqrt{5}}{5}-2\approx -0.211145618$$

$$a=-\frac{4\sqrt{5}}{5}-2\approx -3.788854382$$

Steps Using the Quadratic Formula

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

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

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

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

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$16+16a+5a^{2}+4a+4=16$$

Combine 16a and 4a to get 20a.

$$16+20a+5a^{2}+4=16$$

Add 16 and 4 to get 20.

$$20+20a+5a^{2}=16$$

Subtract 16 from both sides.

$$20+20a+5a^{2}-16=0$$

Subtract 16 from 20 to get 4.

$$4+20a+5a^{2}=0$$

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.

$$5a^{2}+20a+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 20 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-20±\sqrt{20^{2}-4\times 5\times 4}}{2\times 5}$$

Square 20.

$$a=\frac{-20±\sqrt{400-4\times 5\times 4}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-20±\sqrt{400-20\times 4}}{2\times 5}$$

Multiply -20 times 4.

$$a=\frac{-20±\sqrt{400-80}}{2\times 5}$$

Add 400 to -80.

$$a=\frac{-20±\sqrt{320}}{2\times 5}$$

Take the square root of 320.

$$a=\frac{-20±8\sqrt{5}}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{-20±8\sqrt{5}}{10}$$

Now solve the equation a=\frac{-20±8\sqrt{5}}{10} when ± is plus. Add -20 to 8\sqrt{5}.

$$a=\frac{8\sqrt{5}-20}{10}$$

Divide -20+8\sqrt{5} by 10.

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

Now solve the equation a=\frac{-20±8\sqrt{5}}{10} when ± is minus. Subtract 8\sqrt{5} from -20.

$$a=\frac{-8\sqrt{5}-20}{10}$$

Divide -20-8\sqrt{5} by 10.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

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

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

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$16+16a+5a^{2}+4a+4=16$$

Combine 16a and 4a to get 20a.

$$16+20a+5a^{2}+4=16$$

Add 16 and 4 to get 20.

$$20+20a+5a^{2}=16$$

Subtract 20 from both sides.

$$20a+5a^{2}=16-20$$

Subtract 20 from 16 to get -4.

$$20a+5a^{2}=-4$$

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.

$$5a^{2}+20a=-4$$

Divide both sides by 5.

$$\frac{5a^{2}+20a}{5}=-\frac{4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}+\frac{20}{5}a=-\frac{4}{5}$$

Divide 20 by 5.

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

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.

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

Square 2.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 2 from both sides of the equation.

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