Solve sqrt{x^2+2x+1}=x+5

$$\sqrt{ { x }^{ 2 } +2x+1 } =x+5$$

Solve for x

$$x=-3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$x^{2}+2x+1=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$x^{2}+2x+1-x^{2}=10x+25$$

Combine x^{2} and -x^{2} to get 0.

$$2x+1=10x+25$$

Subtract 10x from both sides.

$$2x+1-10x=25$$

Combine 2x and -10x to get -8x.

$$-8x+1=25$$

Subtract 1 from both sides.

$$-8x=25-1$$

Subtract 1 from 25 to get 24.

$$-8x=24$$

Divide both sides by -8.

$$x=\frac{24}{-8}$$

Divide 24 by -8 to get -3.

$$x=-3$$

Substitute -3 for x in the equation \sqrt{x^{2}+2x+1}=x+5.

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

Simplify. The value x=-3 satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}+2x+1}=x+5 has a unique solution.

$$x=-3$$