Solve sqrt{x^2-4}+sqrt{x-2}=0

$$\sqrt { x ^ { 2 } - 4 } + \sqrt { x - 2 } = 0$$

Solve for x

$$x=2$$

Solution Steps

Subtract \sqrt{x-2} from both sides of the equation.

$$\sqrt{x^{2}-4}=-\sqrt{x-2}$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}-4}\right)^{2}=\left(-\sqrt{x-2}\right)^{2}$$

Calculate \sqrt{x^{2}-4} to the power of 2 and get x^{2}-4.

$$x^{2}-4=\left(-\sqrt{x-2}\right)^{2}$$

Expand \left(-\sqrt{x-2}\right)^{2}.

$$x^{2}-4=\left(-1\right)^{2}\left(\sqrt{x-2}\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$x^{2}-4=1\left(\sqrt{x-2}\right)^{2}$$

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

$$x^{2}-4=1\left(x-2\right)$$

Use the distributive property to multiply 1 by x-2.

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

Subtract x from both sides.

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

Add 2 to both sides.

$$x^{2}-4-x+2=0$$

Add -4 and 2 to get -2.

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

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

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

To solve the equation, factor x^{2}-x-2 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.

$$a+b=-1 ab=-2$$

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. The only such pair is the system solution.

$$a=-2 b=1$$

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

$$\left(x-2\right)\left(x+1\right)$$

To find equation solutions, solve x-2=0 and x+1=0.

$$x=2 x=-1$$

Substitute 2 for x in the equation \sqrt{x^{2}-4}+\sqrt{x-2}=0.

$$\sqrt{2^{2}-4}+\sqrt{2-2}=0$$

Simplify. The value x=2 satisfies the equation.

$$0=0$$

Substitute -1 for x in the equation \sqrt{x^{2}-4}+\sqrt{x-2}=0. The expression \sqrt{\left(-1\right)^{2}-4} is undefined because the radicand cannot be negative.

$$\sqrt{\left(-1\right)^{2}-4}+\sqrt{-1-2}=0$$

Equation \sqrt{x^{2}-4}=-\sqrt{x-2} has a unique solution.

$$x=2$$