Solve (x+2)^2+x^2=81

$$( x + 2 ) ^ { 2 } + x ^ { 2 } = 81$$

Solve for x

$$x=\frac{\sqrt{158}}{2}-1\approx 5.284902545$$

$$x=-\frac{\sqrt{158}}{2}-1\approx -7.284902545$$

Steps Using the Quadratic Formula

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

$$x^{2}+4x+4+x^{2}=81$$

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

$$2x^{2}+4x+4=81$$

Subtract 81 from both sides.

$$2x^{2}+4x+4-81=0$$

Subtract 81 from 4 to get -77.

$$2x^{2}+4x-77=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-77\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-77\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-77\right)}}{2\times 2}$$

Multiply -8 times -77.

$$x=\frac{-4±\sqrt{16+616}}{2\times 2}$$

Add 16 to 616.

$$x=\frac{-4±\sqrt{632}}{2\times 2}$$

Take the square root of 632.

$$x=\frac{-4±2\sqrt{158}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±2\sqrt{158}}{4}$$

Now solve the equation x=\frac{-4±2\sqrt{158}}{4} when ± is plus. Add -4 to 2\sqrt{158}.

$$x=\frac{2\sqrt{158}-4}{4}$$

Divide -4+2\sqrt{158} by 4.

$$x=\frac{\sqrt{158}}{2}-1$$

Now solve the equation x=\frac{-4±2\sqrt{158}}{4} when ± is minus. Subtract 2\sqrt{158} from -4.

$$x=\frac{-2\sqrt{158}-4}{4}$$

Divide -4-2\sqrt{158} by 4.

$$x=-\frac{\sqrt{158}}{2}-1$$

The equation is now solved.

$$x=\frac{\sqrt{158}}{2}-1 x=-\frac{\sqrt{158}}{2}-1$$

Steps for Completing the Square

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

$$x^{2}+4x+4+x^{2}=81$$

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

$$2x^{2}+4x+4=81$$

Subtract 4 from both sides.

$$2x^{2}+4x=81-4$$

Subtract 4 from 81 to get 77.

$$2x^{2}+4x=77$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{77}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{77}{2}$$

Divide 4 by 2.

$$x^{2}+2x=\frac{77}{2}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{77}{2}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{77}{2}+1$$

Add \frac{77}{2} to 1.

$$x^{2}+2x+1=\frac{79}{2}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{79}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{79}{2}}$$

Simplify.

$$x+1=\frac{\sqrt{158}}{2} x+1=-\frac{\sqrt{158}}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{158}}{2}-1 x=-\frac{\sqrt{158}}{2}-1$$