Exercises: Terms with Square Roots

What is the maximum domain of definition? If possible, write the term without a square root sign.

$\sqrt{49 a^{4} b^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{49 a^{4} b^{2}}$
This expression is not defined whenever there is a negative number under the root.
Both exponents of $a$ and $b$ are even, so $a^{4}$ and $b^{2}$ are always positive or zero. Thus, the expression under the root is also positive or zero and the root is always defined:
$D_{\max} = ℝ$
Now, take the root.
$\sqrt{49 a^{4} b^{2}} = 7 \cdot a^{2} \cdot | b |$
You have to put dashes, since $b$ could be negative.
$\sqrt{{(- b)}^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{{(- b)}^{2}}$
The root is net defined, if the expression under the root is negative. Now, squares of any number (also of $- b$ ) are always positive of zero, so there are never any problems with the definition of the root.
$D_{\max} = ℝ$
Square $- b$ .
$\sqrt{{(- b)}^{2}}$ $=$ $\sqrt{b^{2}}$
↓
Draw the root so that the square drops out.
$=$ $| b |$
${\sqrt{- b}}^{2}$

For this task you need the following basic knowledge: Square Roots
${\sqrt{- b}}^{2}$
The number under the root must not be negative. $\Rightarrow$ Only negative numbers or 0 are allowed for $b$ .
$D_{\max} = ℝ_{0}^{-}$
${\sqrt{- b}}^{2}$ $= - b$
So finally, power and root cancel.
$\sqrt{{(1 - 2 x)}^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{{(1 - 2 x)}^{2}}$
The number under the root must not be negative. However, the term in the squared bracket is always positive or zero, so we never have any problem with taking the root.
$D_{\max} = ℝ$
$\sqrt{{(1 - 2 x)}^{2}} = | 1 - 2 x |$
Take the root. Root and square cancel each other out, but since the term in the bracket can also be negative and could be made positive by squaring, you have to put an absolute value.
$\sqrt{{(x - y)}^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{{(x - y)}^{2}}$
The number under the root must not be negative. However, squaring makes a term non-negative, so there is never a problem wit the root being defined.
$D_{\max} = ℝ$
$\sqrt{{(x - y)}^{2}} = | x - y |$
Take the root. Root and square cancel each other out, but since the term in the bracket can also be negative and could be made positive by squaring, you have to put an absolute value.
$\sqrt{x^{2} + y^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{x^{2} + y^{2}}$
The number under the root must not be negative. But since both squares are positive or zero, we never get any definition problems with the root.
$D_{\max} = ℝ$
$\sqrt{x^{2} + y^{2}} = \sqrt{x^{2} + y^{2}}$
It is not possible to further simplify this term. It is already in the simplest possible form.
$\sqrt{x^{2} \cdot y^{2}}$

For this task you need the following basic knowledge: Square Roots
$\sqrt{x^{2} \cdot y^{2}}$
Both $x^{2}$ and $y^{2}$ are never zero. The same holds for the product, so there are never any problems with definition of the square root.
$\to 𝔻 = ℝ$
Now, $\sqrt{x^{2} \cdot y^{2}} = \sqrt{{(x \cdot y)}^{2}}$ . Take the root. Root and square cancel each other out, but it is necessary to retain absolute values, since $x$ or $y$ might be negative and turned positive by the squaring.
$= | x \cdot y |$

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$\sqrt{{(- b)}^{2}}$	$=$	$\sqrt{b^{2}}$
	↓	Draw the root so that the square drops out.
	$=$	$\| b \|$