Exercises: Terms with Square Roots

Here you will find mixed exercises on terms with square roots. Learn to simplify square roots and determine the numerical values of roots!

For which numbers are the terms defined? Write them without square roots and think about whether an absolute value is necessary or not.

$5 \cdot \sqrt{2 x} \cdot \sqrt{18 x}$

$\frac{\sqrt{2 a^{2}} \cdot \sqrt{12 a}}{\sqrt{8 a}}$

${(\sqrt{d - 2})}^{2}$

For this task you need the following basic knowledge: Square Roots
Determine the domain of definition
First determine the domain of definition.
${(\sqrt{d - 2})}^{2}$
Only positive expressions are allowed under the root. This gives you the following condition:
$d - 2 \geq 0$
$d \geq 2$
$D = [2; \infty)$
Write the term without square roots
Take the square root and do not forget the absolute value.
${(\sqrt{d - 2})}^{2}$ $=$ $| d - 2 |$
$=$ $d - 2$
$\sqrt{{(d - 2)}^{2}}$

For this task you need the following basic knowledge: Square Roots
Determine the domain of definition
First determine the domain of definition.
$\sqrt{{(d - 2)}^{2}}$
Only positive expressions are allowed under the root. Since the term under the root is squared, it will always be positive. So there are no definition problems, no matter which values you use for $d$ .
$D = ℝ$
Write the term without square roots
$\sqrt{{(d - 2)}^{2}}$ $=$ $| d - 2 |$
↓
Consider whether you can omit the absolute value.
$=$ $| d - 2 |$

2
What is the domain of definition?
$\sqrt{x - 36}$

For this task you need the following basic knowledge: Square Roots
Only positive or zero expressions are allowed under the root. So $x - 36$ must be positive or zero. $x - 36$ becomes zero for $x = 36$ .
When we make $x$ smaller (so $x < 36$ ), then $x - 36$ gets negative (which is not good)
When we make $x$ larger (so $x > 36$ ), then $x - 36$ becomes positive (which is what we want).
So $x \geq 36$ is the required condition for obtaining a positive or zero number under the root. Further, $x$ is a real number, so
$x \in ℝ, x ⩾ 36$

$\sqrt{36 + x^{2}}$

For this task you need the following basic knowledge: Square Roots
Only positive or zero expressions are allowed under the root. So $36 + x^{2}$ must not be negative. Now, $x^{2}$ is never negative and if we add $+ 36$ , it only becomes "more positive". So any $x$ is allowed in the domain of definition, as long as it is a real number:
$D = ℝ$

$\frac{1}{\sqrt{x + 36}}$

For this task you need the following basic knowledge: Square Roots
Only positive or zero expressions are allowed under the root. So $x + 36$ must not be negative. Further, we cannot divide by zero, which excludes the case $\sqrt{x + 36} = 0$ , which happens if $x + 36$ becomes zero. So only $x + 36 > 0$ is allowed. This is the case whenever $x > - 36$ . Further, we know that $x$ is a real number.
So the domain of definition of $\frac{1}{\sqrt{x + 36}}$ is characterized by
$x \in R, x > - 36$

$\sqrt{x^{2} - 36}$

For this task you need the following basic knowledge: Square Roots
Only positive or zero expressions are allowed under the root. So $x^{2} - 36$ must not be negative. That means, $x^{2}$ must be larger or equal $36$ .
For $x = 6$ and $x = - 6$ , you exactly obtain $x^{2} = 36$ . This is a "good case"
If $x$ is positive, then making $x$ larger (so $x > 6$ ) will make $x^{2}$ larger than $36$ , which is also a "good case".
Analogously, for negative $x$ , making $x$ even more negative (so $x < - 6$ ) will also result in $x^{2} > 36$ , which is good for our purposes.
However, if - $6 < x < 6$ , so $x \in (- 6, 6)$ , then $x^{2}$ becomes smaller than 36, which is not good.. So we have to exclude this case.
Further, we know that $x$ is a real number. So the domain of definition is given by
$D = ℝ \ (- 6,6)$
3
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.
$= | x \cdot y |$
4
Simplify the following terms.
$\sqrt{500} + 3 \sqrt{98} - 5 \sqrt{8} - 3 \sqrt{45}$

For this task you need the following basic knowledge: Square Roots
$=$ $\sqrt{500} + 3 \sqrt{98} - 5 \sqrt{8} - 3 \sqrt{45}$
↓
Decompose the numbers under the roots.
$=$ $\sqrt{5 \cdot 10^{2}} + 3 \sqrt{2 \cdot 7^{2}} - 5 \sqrt{2^{2} \cdot 2} - 3 \sqrt{3^{2} \cdot 5}$
↓
Pull the powers out of the root.
$=$ $10 \sqrt{5} + 3 \cdot 7 \sqrt{2} - 5 \cdot 2 \sqrt{2} - 3 \cdot 3 \sqrt{5}$
$=$ $10 \sqrt{5} + 21 \sqrt{2} - 10 \sqrt{2} - 9 \sqrt{5}$
$=$ $\sqrt{5} + 11 \sqrt{2}$

$\sqrt{64 k^{2}}$

For this task you need the following basic knowledge: Square Roots
$=$ $\sqrt{64 k^{2}}$
↓
Decompose the numbers under the root.
$=$ $\sqrt{8^{2} \cdot k^{2}}$
↓
Pull the powers out of the root.
(Don't forget the absolute value around k.)
$=$ $8 \cdot | k |$

$(\sqrt{\frac{x^{5} y}{5 a}} : \sqrt{\frac{x^{3} y^{3}}{a^{2}}}) \cdot \sqrt{\frac{25 x}{a}} (x, y, z > 0)$

For this task you need the following basic knowledge: Square Roots
$=$ $(\sqrt{\frac{x^{5} y}{5 a}} : \sqrt{\frac{x^{3} y^{3}}{a^{2}}}) \cdot \sqrt{\frac{25 x}{a}}$
↓
Conclude the fractions.
$=$ $\frac{\sqrt{x^{5} y} \cdot \sqrt{a^{2}}}{\sqrt{5 a} \cdot \sqrt{x^{3} y^{3}}} \cdot \frac{\sqrt{25 x}}{\sqrt{a}}$
$=$ $\sqrt{\frac{x^{5} y \cdot a^{2} \cdot 25 x}{5 a x^{3} y^{3} a}}$
↓
Shorten the fraction.
$=$ $\sqrt{\frac{5 x^{3}}{y^{2}}}$
↓
Pull the powers out of the root.
$=$ $\frac{x}{y} \sqrt{5 x}$
5
Simplify the term $\sqrt{169} \cdot x + \sqrt{169 \cdot x^{2}}$ and determine for which values of $x$ the term gets $0$ .

For this task you need the following basic knowledge: 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.
$\sqrt{169} \cdot x + \sqrt{169 \cdot x^{2}}$ $=$ $\sqrt{13^{2}} \cdot x + \sqrt{13^{2} \cdot x^{2}}$
↓
Use the calculation rules for square roots and absolute values.
$=$ $13 \cdot x + 13 \cdot | x |$
Resolve the absolute value
Now you have to solve the absolute value. For this you need a case distinction.
Case 1: $x \geq 0$
If one only uses positive $x$ -values, you can also omit the dashes.
$13 \cdot x + 13 \cdot | x |$ $=$ $13 x + 13 x$
$=$ $26 x$
$26 x = 0$ holds if and only if $x = 0$ .
Case 1: $x < 0$
$13 \cdot x + 13 \cdot | x |$
If you only use negative $x$ -values, you can plug in $- x$ for $| x |$
$13 \cdot x + 13 \cdot | x |$ $=$ $13 x + 13 \cdot (- x)$
$=$ $13 x - 13 x$
$=$ $0$
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 $x \leq 0$ , i.e., for $x = 0$ and all negative numbers.

When solving quadratic equations, one obtains expressions with square roots, like the following two. Simplify them.

$x_{1 / 2} = \frac{- 5 \pm \sqrt{5^{2} + 4 \cdot \sqrt{7} \cdot 2 \sqrt{7}}}{2 \sqrt{7}}$

For this task you need the following basic knowledge: Square Roots
$x_{1 / 2}$ $=$ $\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot 8}}{2}$
↓
Conclude everything under the root.
$=$ $\frac{- 14 \pm \sqrt{164}}{2}$
↓
Pull 2 out of the root.
$=$ $\frac{- 14 \pm \sqrt{4 \cdot 41}}{2} = \frac{- 14 \pm 2 \sqrt{41}}{2}$
↓
Shorten the fractions.
$=$ $- 7 \pm \sqrt{41}$

$x_{1 / 2} = \frac{- 5 \pm \sqrt{5^{2} + 4 \cdot \sqrt{7} \cdot 2 \sqrt{7}}}{2 \sqrt{7}}$

7
Give an argumentation, for why $\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$ for all positive $a$ .

For this task you need the following basic knowledge: Square Roots
We would like to show that $\frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{a}$
Try to make the denominator rational for this.
$\frac{1}{\sqrt{a}}$
Expand by $\sqrt{a}$
$= \frac{1 \cdot \sqrt{a}}{\sqrt{a} \cdot \sqrt{a}}$
Use the calculation rules for the product of roots.
$= \frac{1 \cdot \sqrt{a}}{\sqrt{a \cdot a}}$
The square and the root cancel each other.
$= \frac{\sqrt{a}}{a}$
8
For which values of $x$ are the following statements true?
$\sqrt{x^{2}} = - x$

For this task you need the following basic knowledge: Square Roots
For which $x$ do we have $\sqrt{x^{2}} = - x$ ?
In general, we have $\sqrt{x^{2}} = | x |$ . So:
For which $x$ is it true that $| x | = - x$ ?
Answer: For all values of $x$ that are less than or equal to $0$ .

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

For this task you need the following basic knowledge: Square Roots
This is true for all values $x \geq 1$ , since $\sqrt{{(x - 1)}^{2}} = | x - 1 |$ and the absolute value function is equal to the right-hand side exactly for $x - 1 \geq 0$ .

$x_{1 / 2}$	$=$	$\frac{- 5 \pm \sqrt{5^{2} + 4 \cdot \sqrt{7} \cdot 2 \sqrt{7}}}{2 \sqrt{7}}$
	↓	Conclude everything under the root.
$x_{1 / 2}$	$=$	$\frac{- 5 \pm \sqrt{81}}{2 \sqrt{7}} = \frac{- 5 \pm 9}{2 \sqrt{7}}$
$x_{1}$	$=$	$\frac{- 5 + 9}{2 \sqrt{7}} = \frac{4}{2 \sqrt{7}}$
	↓	Shorten by 2
$x_{1}$	$=$	$\frac{2}{\sqrt{7}}$
	↓	Expand by $\sqrt{7}$ .
$x_{1}$	$=$	$\frac{2 \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} = \frac{2 \sqrt{7}}{7}$
$x_{2}$	$=$	$\frac{- 5 - 9}{2 \sqrt{7}} = \frac{- 14}{2 \sqrt{7}}$
	↓	Expand by $\sqrt{7}$ .
$x_{2}$	$=$	$\frac{- 14 \sqrt{7}}{2 \sqrt{7} \sqrt{7}} = \frac{- 14 \sqrt{7}}{2 \cdot 7} = \frac{- 14 \sqrt{7}}{14}$
	↓	Shorten by 14.
$x_{2}$	$=$	$- \sqrt{7}$

$5 \cdot \sqrt{2 x} \cdot \sqrt{18 x}$	$=$	$5 \cdot \sqrt{2 x \cdot 18 x}$
	↓	Conclude everything below the root.
	$=$	$5 \cdot \sqrt{36 x^{2}}$
	↓	Take the square root and do not forget the absolute value.
	$=$	$5 \cdot 6 \cdot \| x \|$
	↓	Conclude and consider whether you can omit the absolute value.
	$=$	$30 x$

$\frac{\sqrt{2 a^{2}} \cdot \sqrt{12 a}}{\sqrt{8 a}}$	$=$	$\sqrt{\frac{2 a^{2} \cdot 12 a}{8 a}}$
	↓	Conclude everything below the root.
	$=$	$\sqrt{3 a^{2}}$
	↓	Take the square root and do not forget the absolute value.
	$=$	$\sqrt{3} \cdot \| a \|$
	↓	Conclude and consider whether you can omit the absolute value.
	$=$	$\sqrt{3} \cdot a$

$\sqrt{{(d - 2)}^{2}}$	$=$	$\| d - 2 \|$
	↓	Consider whether you can omit the absolute value.
	$=$	$\| d - 2 \|$

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

	$=$	$\sqrt{500} + 3 \sqrt{98} - 5 \sqrt{8} - 3 \sqrt{45}$
	↓	Decompose the numbers under the roots.
	$=$	$\sqrt{5 \cdot 10^{2}} + 3 \sqrt{2 \cdot 7^{2}} - 5 \sqrt{2^{2} \cdot 2} - 3 \sqrt{3^{2} \cdot 5}$
	↓	Pull the powers out of the root.
	$=$	$10 \sqrt{5} + 3 \cdot 7 \sqrt{2} - 5 \cdot 2 \sqrt{2} - 3 \cdot 3 \sqrt{5}$
	$=$	$10 \sqrt{5} + 21 \sqrt{2} - 10 \sqrt{2} - 9 \sqrt{5}$
	$=$	$\sqrt{5} + 11 \sqrt{2}$

	$=$	$\sqrt{64 k^{2}}$
	↓	Decompose the numbers under the root.
	$=$	$\sqrt{8^{2} \cdot k^{2}}$
	↓	Pull the powers out of the root. (Don't forget the absolute value around k.)
	$=$	$8 \cdot \| k \|$

	$=$	$(\sqrt{\frac{x^{5} y}{5 a}} : \sqrt{\frac{x^{3} y^{3}}{a^{2}}}) \cdot \sqrt{\frac{25 x}{a}}$
	↓	Conclude the fractions.
	$=$	$\frac{\sqrt{x^{5} y} \cdot \sqrt{a^{2}}}{\sqrt{5 a} \cdot \sqrt{x^{3} y^{3}}} \cdot \frac{\sqrt{25 x}}{\sqrt{a}}$
	$=$	$\sqrt{\frac{x^{5} y \cdot a^{2} \cdot 25 x}{5 a x^{3} y^{3} a}}$
	↓	Shorten the fraction.
	$=$	$\sqrt{\frac{5 x^{3}}{y^{2}}}$
	↓	Pull the powers out of the root.
	$=$	$\frac{x}{y} \sqrt{5 x}$

$\sqrt{169} \cdot x + \sqrt{169 \cdot x^{2}}$	$=$	$\sqrt{13^{2}} \cdot x + \sqrt{13^{2} \cdot x^{2}}$
	↓	Use the calculation rules for square roots and absolute values.
	$=$	$13 \cdot x + 13 \cdot \| x \|$

$13 \cdot x + 13 \cdot \| x \|$	$=$	$13 x + 13 \cdot (- x)$
	$=$	$13 x - 13 x$
	$=$	$0$

$x_{1 / 2}$	$=$	$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot 8}}{2}$
	↓	Conclude everything under the root.
	$=$	$\frac{- 14 \pm \sqrt{164}}{2}$
	↓	Pull 2 out of the root.
	$=$	$\frac{- 14 \pm \sqrt{4 \cdot 41}}{2} = \frac{- 14 \pm 2 \sqrt{41}}{2}$
	↓	Shorten the fractions.
	$=$	$- 7 \pm \sqrt{41}$

${(\sqrt{d - 2})}^{2}$	$=$	$\| d - 2 \|$
	$=$	$d - 2$

Exercises: Terms with Square Roots

Determine the domain of definition

Write the term without square roots

Determine the domain of definition

Write the term without square roots

Determine the domain of definition

Write the term without square roots

Determine the domain of definition

Write the term without square roots