Exercises: Terms with Square Roots

Determine the domain of definition:

$\sqrt{169}\cdot x+\sqrt{169\cdot x^2}$

For the first term, $x$ stands outside of the root. You may therefore use any real number as $x$.

In the second term, $x^2$ is under the root. The term under the root must not be negative for the root to be defined. Now, $x^2$ is non-negative for any real number $x$, and the same holds for $169\cdot x^2$.

So you can use any real number as $x$.

$D=ℝ$

Compute the root

$\sqrt{169}\cdot x+\sqrt{169\cdot x^2}$

Write 169 as a square number.

Resolve the absolute value

Now you have to solve the absolute value. For this you need a case distinction.

Case 1: $x\ge 0$

If one only uses positive $x$ -values, you can also omit the dashes.

$26x=0$ holds if and only if $x=0$ .

Case 1: $x<0$

$13\cdot x+13\cdot \left|x\right|$

If you only use negative $x$-values, you can plug in $-x$ for $|x|$

For negative $x$-values, this expression is always 0.

Thus, the term $\sqrt{169}\cdot x+\sqrt{169\cdot x^2}$ is $0$ for all $x\le 0$ , i.e., for $x=0$ and all negative numbers.