Precalculus Examples

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

Step 1

Set the radicand in $\sqrt{x - 1}$ greater than or equal to $0$ to find where the expression is defined.

$x - 1 \geq 0$

Step 2

Add $1$ to both sides of the inequality.

$x \geq 1$

Step 3

Set the radicand in $\sqrt{x - 2}$ greater than or equal to $0$ to find where the expression is defined.

$x - 2 \geq 0$

Step 4

Add $2$ to both sides of the inequality.

$x \geq 2$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${x | x \geq 2}$

Step 6