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

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

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

$$3=2x+1$$

Swap sides so that all variable terms are on the left hand side.

$$2x+1=3$$

Subtract 1 from both sides.

$$2x=3-1$$

Subtract 1 from 3 to get 2.

$$2x=2$$

Divide both sides by 2.

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

Divide 2 by 2 to get 1.

$$x=1$$

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

$$\sqrt{1^{2}+3}=1+1$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Equation \sqrt{x^{2}+3}=x+1 has a unique solution.

$$x=1$$