Solve sqrt{(x+1)^2+(2x)^2}=sqrt{(x+3)^2}

$$\sqrt { ( x + 1 ) ^ { 2 } + ( 2 x ) ^ { 2 } } = \sqrt { ( x + 3 ) ^ { 2 } }$$

Solve for x

$$x=2$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$\left(\sqrt{x^{2}+2x+1+4x^{2}}\right)^{2}=\left(\sqrt{\left(x+3\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{5x^{2}+2x+1}\right)^{2}=\left(\sqrt{\left(x+3\right)^{2}}\right)^{2}$$

Calculate \sqrt{5x^{2}+2x+1} to the power of 2 and get 5x^{2}+2x+1.

$$5x^{2}+2x+1=\left(\sqrt{\left(x+3\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$5x^{2}+2x+1=\left(\sqrt{x^{2}+6x+9}\right)^{2}$$

Calculate \sqrt{x^{2}+6x+9} to the power of 2 and get x^{2}+6x+9.

$$5x^{2}+2x+1=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$5x^{2}+2x+1-x^{2}=6x+9$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+2x+1=6x+9$$

Subtract 6x from both sides.

$$4x^{2}+2x+1-6x=9$$

Combine 2x and -6x to get -4x.

$$4x^{2}-4x+1=9$$

Subtract 9 from both sides.

$$4x^{2}-4x+1-9=0$$

Subtract 9 from 1 to get -8.

$$4x^{2}-4x-8=0$$

Divide both sides by 4.

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

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

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

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

$$a=-2 b=1$$

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

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

Factor out x in x^{2}-2x.

$$x\left(x-2\right)+x-2$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Substitute 2 for x in the equation \sqrt{\left(x+1\right)^{2}+\left(2x\right)^{2}}=\sqrt{\left(x+3\right)^{2}}.

$$\sqrt{\left(2+1\right)^{2}+\left(2\times 2\right)^{2}}=\sqrt{\left(2+3\right)^{2}}$$

Simplify. The value x=2 satisfies the equation.

$$5=5$$

Substitute -1 for x in the equation \sqrt{\left(x+1\right)^{2}+\left(2x\right)^{2}}=\sqrt{\left(x+3\right)^{2}}.

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

Simplify. The value x=-1 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{\left(x+1\right)^{2}+\left(2x\right)^{2}}=\sqrt{\left(x+3\right)^{2}}.

$$x=2 x=-1$$