Solve sqrt{x^2+6x+3}-x=2

$$\sqrt { x ^ { 2 } + 6 x + 3 } - x = 2$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{x^{2}+6x+3}=2+x$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}+6x+3}\right)^{2}=\left(2+x\right)^{2}$$

Calculate \sqrt{x^{2}+6x+3} to the power of 2 and get x^{2}+6x+3.

$$x^{2}+6x+3=\left(2+x\right)^{2}$$

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

$$x^{2}+6x+3=4+4x+x^{2}$$

Subtract 4x from both sides.

$$x^{2}+6x+3-4x=4+x^{2}$$

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

$$x^{2}+2x+3=4+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+2x+3-x^{2}=4$$

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

$$2x+3=4$$

Subtract 3 from both sides.

$$2x=4-3$$

Subtract 3 from 4 to get 1.

$$2x=1$$

Divide both sides by 2.

$$x=\frac{1}{2}$$

Substitute \frac{1}{2} for x in the equation \sqrt{x^{2}+6x+3}-x=2.

$$\sqrt{\left(\frac{1}{2}\right)^{2}+6\times \frac{1}{2}+3}-\frac{1}{2}=2$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}+6x+3}=x+2 has a unique solution.

$$x=\frac{1}{2}$$