Solve (5+a)^2+a=8+a

$$( 5 + a ) ^ { 2 } + a = 8 + a$$

Solve for a

$$a=2\sqrt{2}-5\approx -2.171572875$$

$$a=-2\sqrt{2}-5\approx -7.828427125$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+a\right)^{2}.

$$25+10a+a^{2}+a=8+a$$

Combine 10a and a to get 11a.

$$25+11a+a^{2}=8+a$$

Subtract 8 from both sides.

$$25+11a+a^{2}-8=a$$

Subtract 8 from 25 to get 17.

$$17+11a+a^{2}=a$$

Subtract a from both sides.

$$17+11a+a^{2}-a=0$$

Combine 11a and -a to get 10a.

$$17+10a+a^{2}=0$$

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.

$$a^{2}+10a+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-10±\sqrt{10^{2}-4\times 17}}{2}$$

Square 10.

$$a=\frac{-10±\sqrt{100-4\times 17}}{2}$$

Multiply -4 times 17.

$$a=\frac{-10±\sqrt{100-68}}{2}$$

Add 100 to -68.

$$a=\frac{-10±\sqrt{32}}{2}$$

Take the square root of 32.

$$a=\frac{-10±4\sqrt{2}}{2}$$

Now solve the equation a=\frac{-10±4\sqrt{2}}{2} when ± is plus. Add -10 to 4\sqrt{2}.

$$a=\frac{4\sqrt{2}-10}{2}$$

Divide -10+4\sqrt{2} by 2.

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

Now solve the equation a=\frac{-10±4\sqrt{2}}{2} when ± is minus. Subtract 4\sqrt{2} from -10.

$$a=\frac{-4\sqrt{2}-10}{2}$$

Divide -10-4\sqrt{2} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+a\right)^{2}.

$$25+10a+a^{2}+a=8+a$$

Combine 10a and a to get 11a.

$$25+11a+a^{2}=8+a$$

Subtract a from both sides.

$$25+11a+a^{2}-a=8$$

Combine 11a and -a to get 10a.

$$25+10a+a^{2}=8$$

Subtract 25 from both sides.

$$10a+a^{2}=8-25$$

Subtract 25 from 8 to get -17.

$$10a+a^{2}=-17$$

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.

$$a^{2}+10a=-17$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+10a+5^{2}=-17+5^{2}$$

Square 5.

$$a^{2}+10a+25=-17+25$$

Add -17 to 25.

$$a^{2}+10a+25=8$$

Factor a^{2}+10a+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+5\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+5\right)^{2}}=\sqrt{8}$$

Simplify.

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

Subtract 5 from both sides of the equation.

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