Simplifying Square Roots with Variables

In this article, you will learn all the rules applicable to exercises that have roots with variables and how to find the conditions of the letters (or variables) in the radicand.
Does it sound complicated? Don't worry! A simple lesson, some exercises, and you'll shine.

The general property of roots with variables is the domain of definition of the root.

The number or mathematical expression included in the radicand can be $0$ or greater than $0$, but it can never be negative.

That is
$\sqrt X$
The condition: $X≥0$
$X$ must be greater than or equal to $0$.

After understanding this very important property, we will continue with other laws emphasizing the topic of variables.

When there is a root in each product separately, we can break down the factors and apply the root to them, leaving the multiplication sign between them.
In the following way:
$\sqrt{a \times b} = \sqrt a \times \sqrt b$

As we have seen, the number or the entire expression under the radical must be greater than or equal to $0$, therefore, the condition is that:

$a \times b≥0$
Observe:
Since the product rule allows the breaking down of each factor separately and the root can be applied to it, it is not enough that the product of $a$ and $b$ be greater than or equal to $0$, but the condition must affect both $a$ and $b$ by themselves.
Therefore, all the conditions will be seen in the following way:

$a \times b≥0$
$a≥0$
$b≥0$

When the root acts on the entire quotient (the whole equation) we can break down the factors and apply the root to them, leaving the division sign (the fraction bar) between them.
In the following way:

$\sqrt \frac {X}{Y} = \frac{\sqrt X}{\sqrt Y}$

We will base ourselves on the general law that says the number under the root must be greater than or equal to $0$ and we will obtain the following conditions:
$\frac{a}{b} ≥ 0$
Based on the law of the root of a quotient we will obtain:
$a≥0$
$b>0$

Pay attention! Since $b$ is in the denominator, it cannot be $0$, therefore, the condition is only greater than $0$.
Now let's practice!
What is the condition of the variable $X$?

$\sqrt \frac{X+4}{2} =$

Solution:
According to the law of the root of a quotient, the numerator must be greater than or equal to $0$ and the denominator must be greater than $0$.
The denominator is $2$, a positive number.
Now we just need to verify that the numerator is also greater than or equal to $0$.
We will copy the numerator with the condition and obtain:
$x+4≥0$
Transpose the members and we will obtain:
$x≥-4$
This is the condition of the variable $X$.

What is the condition of the variable $X$?
$\sqrt{3 \times X-2} =$
Solution:
We know that the number or the entire expression under the radical must be equal to or greater than $0$. We will copy the expression under the radical with the condition and obtain:
$3 \times X-2≥0$
Transposing terms, we obtain:
$3X≥2$
$X≥2/3$
This is the condition of $X$.

What is the condition of $X$ and $Y$?
$\sqrt \frac {2X-1}{Y+2} =$

What is the condition of $X$ and $Y$?
$\sqrt \frac {2X-1}{Y+2} =$

Solution:
Before starting to solve, let's remember all the conditions that must be met.
The first rule is that the number under the radical sign must be positive or equal to $0$.
That is: $\sqrt \frac {2X-1}{Y+2}$
According to the quotient rule of the root, we can also express the exercise in the following way:

$\frac {\sqrt {2X-1}}{\sqrt {Y+2}} =$
The numerator $2X-1$ is under the root, therefore, it must be greater or equal to $0$.
We will copy the numerator with the condition and obtain:
$2X-1≥0$
Transposing members we will obtain:
$2X≥1$
$x≥1/2$
This is the condition of $X$.
Now let's move to the denominator
If we acted automatically, we would say that the entire expression under the radical must be greater or equal to $0$. But! The detail-oriented surely have noticed that it is the denominator! and, therefore, it cannot be equal to $0$. So, the condition will be that the denominator is greater than $0$.
We will obtain:
$y+2>0$
Transposing members we will obtain:
$Y>-2$
This is the condition of $Y$.