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

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

Solve for x

$$x=2$$

Solution Steps

Subtract x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$x^{2}-3=9-6x+x^{2}$$

Add 6x to both sides.

$$x^{2}-3+6x=9+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-3+6x-x^{2}=9$$

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

$$-3+6x=9$$

Add 3 to both sides.

$$6x=9+3$$

Add 9 and 3 to get 12.

$$6x=12$$

Divide both sides by 6.

$$x=\frac{12}{6}$$

Divide 12 by 6 to get 2.

$$x=2$$

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

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

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

$$x=2$$