Solve sqrt{x^2-3}-sqrt{x+3}=0

$$\sqrt { x ^ { 2 } - 3 } - \sqrt { x + 3 } = 0$$

Solve for x

$$x=-2$$

$$x=3$$

Solution Steps

Subtract -\sqrt{x+3} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract x from both sides.

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

Subtract 3 from both sides.

$$x^{2}-3-x-3=0$$

Subtract 3 from -3 to get -6.

$$x^{2}-6-x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-x-6=0$$

To solve the equation, factor x^{2}-x-6 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6$$

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. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Substitute 3 for x in the equation \sqrt{x^{2}-3}-\sqrt{x+3}=0.

$$\sqrt{3^{2}-3}-\sqrt{3+3}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute -2 for x in the equation \sqrt{x^{2}-3}-\sqrt{x+3}=0.

$$\sqrt{\left(-2\right)^{2}-3}-\sqrt{-2+3}=0$$

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x^{2}-3}=\sqrt{x+3}.

$$x=3 x=-2$$