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

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

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

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

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

Square 4.

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

Multiply -4 times -2.

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

Multiply 8 times 4.

$$x=\frac{-4±\sqrt{16+32}}{2\left(-2\right)}$$

Add 16 to 32.

$$x=\frac{-4±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

$$x=\frac{-4±4\sqrt{3}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-4±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{-4±4\sqrt{3}}{-4} when ± is plus. Add -4 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-4}{-4}$$

Divide -4+4\sqrt{3} by -4.

$$x=1-\sqrt{3}$$

Now solve the equation x=\frac{-4±4\sqrt{3}}{-4} when ± is minus. Subtract 4\sqrt{3} from -4.

$$x=\frac{-4\sqrt{3}-4}{-4}$$

Divide -4-4\sqrt{3} by -4.

$$x=\sqrt{3}+1$$

The equation is now solved.

$$x=1-\sqrt{3} x=\sqrt{3}+1$$

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}.

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

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

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

Combine 2x^{2} and -4x^{2} to get -2x^{2}.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

$$x^{2}-2x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x^{2}-2x=2$$

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.

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

Add 2 to 1.

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

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}.

$$\left(x-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{3}$$

Simplify.

$$x-1=\sqrt{3} x-1=-\sqrt{3}$$

Add 1 to both sides of the equation.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$