Solve (x+2)^2+x^2=(sqrt{10})^2

$$( x + 2 ) ^ { 2 } + x ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$$

Solve for x

$$x=-3$$

$$x=1$$

Steps Using Factoring By Grouping

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

$$x^{2}+4x+4+x^{2}=\left(\sqrt{10}\right)^{2}$$

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

$$2x^{2}+4x+4=\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$2x^{2}+4x+4=10$$

Subtract 10 from both sides.

$$2x^{2}+4x+4-10=0$$

Subtract 10 from 4 to get -6.

$$2x^{2}+4x-6=0$$

Divide both sides by 2.

$$x^{2}+2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).

$$\left(x^{2}-x\right)+\left(3x-3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

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}.

$$x^{2}+4x+4+x^{2}=\left(\sqrt{10}\right)^{2}$$

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

$$2x^{2}+4x+4=\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$2x^{2}+4x+4=10$$

Subtract 10 from both sides.

$$2x^{2}+4x+4-10=0$$

Subtract 10 from 4 to get -6.

$$2x^{2}+4x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-4±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

$$x=\frac{-4±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$x=\frac{-4±8}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±8}{4}$$

Now solve the equation x=\frac{-4±8}{4} when ± is plus. Add -4 to 8.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{-4±8}{4} when ± is minus. Subtract 8 from -4.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

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}.

$$x^{2}+4x+4+x^{2}=\left(\sqrt{10}\right)^{2}$$

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

$$2x^{2}+4x+4=\left(\sqrt{10}\right)^{2}$$

The square of \sqrt{10} is 10.

$$2x^{2}+4x+4=10$$

Subtract 4 from both sides.

$$2x^{2}+4x=10-4$$

Subtract 4 from 10 to get 6.

$$2x^{2}+4x=6$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{6}{2}$$

Divide 4 by 2.

$$x^{2}+2x=\frac{6}{2}$$

Divide 6 by 2.

$$x^{2}+2x=3$$

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.

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$