Square Roots

The Square Root

For some given positive number $a$ , the square root $b = \sqrt{a}$ is that number with $b^2=(\sqrt{a})^2=a$ . So when you square the square root $b$ , then it becomes again $a$ .

The square root $b=\sqrt{a}$ is always positive or $0$ .

BeachteWhy is the square root always a positive number?

Looking at the example $b = \sqrt{4}$ , the square root is the number you have to square to get $a = 4$ . Obviously, $b = 2$ solves the problem. But $b = - 2$ would also be a reasonable solution, because $(-2)\cdot(-2)\cdot=(-2)^2=4$ .

So why is (-2) not the square root of 4?

This is because we want the square root to be uniquely defined. If $\sqrt 4$ could mean both $- 2$ and $2$ , then the statement $\sqrt 4 = 2$ would be both true and false, which does not make any sense. So we only restrict to the positive value $\sqrt{4} = 2$ and the statement is true.

Another motivation is that it is useful to define the square root as a function $f(x) = \sqrt{x}$ . A function $f$ assigns to each $x$ in its domain exactly one number $f (x)$ . So if we would like $f(x) = \sqrt{x}$ to be a function, then $\sqrt{x}$ has to be uniquely defined.

The restriction to the positive value is a matter of convention. In principle, one could also take the negative value, but taking the positive value is easier since we have to worry less about minus signs. So we take the positive value.

The square root of $a$ is sometimes also denoted as $\sqrt[2]{a}$ . This is, because also higher-order roots $\sqrt[3]{a}$ , $\sqrt[4]{a}$ , ... exist (these are NOT square roots). The numbers $3, 4, \ldots$ are also called indices, here.

The number $a$ is sometimes also called radicand. It is always positive or $0$ .

Examples

$\sqrt{4}=2$ , because $2^{2} = 4$ . Careful: $(-2)\times(-2)=4$ , but $- 2$ is not the solution, because the square root of a number is always positive.
$\sqrt9=3$ , because $3^{2} = 9$ .
$\sqrt{81}=9$ , because $9^{2} = 81$ .
$\sqrt{-3}$ does not exist, because the radicand is negative.

Square roots of terms

You can not only take roots of numbers, but also of terms. Also here, the radicand (= the term under the root) must not become negative. And just as with square roots of numbers, the square root of a term is always positive or 0.

Examples

$\sqrt{5x+8}$
$\sqrt{(a+2)^2}$
$-\sqrt{x-7}$

Domain of definition

When taking roots from terms, you have to make sure that the radicand is not negative. You can do this by carefully adjusting your domain of definition.

Square roots and absolute values

If there is a square term $x^{2}$ under the square root $\sqrt{x^2}$ , and you want to resolve the root, then you have to take the absolute value:

$\sqrt{x^2}=\left|x\right|$

This is because square roots must be positive and you want to obtain a positive result.

BeachteWhat if you drop the absloute value?

If you would only write $\sqrt{x^2}=x$ , then the result would be wrong for negative numbers of $x$ .

As an example, for $x = - 3$ , we would have $\sqrt{x^2} = \sqrt{(-3)^2} = \sqrt{9} = 3$ , which is not equal to $- 3$ .

The reason is that the squaring "deletes the information" whether $x$ is positive or negative: Both $x = 3$ and $x = - 3$ result in $x^{2} = 9$ . This information remains lost in $\sqrt{x^2}$ . Also taking the absolute value deletes this information, since both $x = 3$ and $x = - 3$ results in $∣ x ∣ = 3$ . So after deleting the information on both sides, the equation $\sqrt{x^2}=\left|x\right|$ is correct.

$\left( \sqrt{x} \right)^2=x$

BeachteWarum kann man hier die Betragsstriche weglassen?

$\sqrt{x}$ does only exist if $x$ is positive or $0$ . Negative $x$ are always excluded, so taking $(\sqrt{x})^2$ does not delete any further information. Therefore, only putting $\left( \sqrt{x} \right)^2=x$ is correct.

However, since negative $x$ are excluded, also taking the absolute value $∣ x ∣$ does not delete any information. Therefore, it is also correct to write $\left(\sqrt x\right)^2=\left|x\right|$ .

How to handle roots of terms

Generally	Example $\sqrt{6x^2}$
1. First determine the domain of definition for the radicand.	The radicand is $6 x^{2}$ . It never becomes negative because the variable $x$ is squared. Therefore, the domain of definition is all of $\mathbb{R}$ , i.e. all positive and negative numbers.
2. Take the root and then an absolute value.	$\sqrt{6x^2} = \left\|\sqrt{6}\cdot x\right\|$
Consider whether you can drop the absolute value. It can be drop if the term inside the absolute value is positive or 0 for all numbers in the domain of definition.	If one were to use negative values for $x$ (these are in the domain of definition), one would get the expression $-\sqrt6\cdot x$ in the absolute value, which is not $\sqrt{6} \cdot x$ . So you must not drop the absolute value.

Calculation rules

Calculation rule	Example
$\sqrt a\;\cdot\;\sqrt{\mathrm b}=\sqrt[{}]{\mathrm{ab}}$ $\frac{\sqrt a}{\sqrt b}=\sqrt{\frac {a}{b}}$ $\sqrt{a^2}=\left\|a\right\|$ $\left( \sqrt {a} \right)^2= a$ $\sqrt a\cdot\sqrt a=a$ $\sqrt{a}=a^{\frac{1}{2}}$	$\sqrt{12}=\sqrt{4\cdot3}=\sqrt4\cdot\sqrt3=2\sqrt3$ $\frac{\sqrt{36}}{\sqrt4}=\sqrt{\frac{36}4}=\sqrt9=3$ $\sqrt{\left(-2\right)^2}=\left\|-2\right\|=2$ $\left(\sqrt {21} \right)^2=21$ $\sqrt3\cdot\sqrt3=3$ $\sqrt{17}=17^{\frac{1}{2}}$

Making denominators rational

If a number is given by $\dfrac{a}{\sqrt{b}}$ , then you can expand this fraction with $\sqrt b$ to eliminate the root from the denominator. The calculation reads as follows:

$\dfrac a{\sqrt b}=\dfrac a{\sqrt b}\cdot\dfrac{\sqrt b}{\sqrt b}=\dfrac{a\cdot\sqrt b}{\sqrt b\cdot\sqrt b}=\dfrac{a\sqrt b}b$

Taking square roots in equations

If you use roots to simplify equations, you have to be careful not to lose some solutions by accidentally "deleting information"! That's why you have to use the absolute value here too.

A simple example will illustrate this:

$\displaystyle x^2$	$=$	$\displaystyle 4$	$\displaystyle \sqrt{}$
	↓	Take the root on both sides
$\displaystyle \sqrt{x^2}$	$=$	$\displaystyle \sqrt{4}$
	↓	Now take the root on both sides according to the above calculation rules (including absolute values).
$\displaystyle \|x\|$	$=$	$\displaystyle 2$
	↓	Here it is important not to forget the absloute value. If you resolve the absolute value, you get two sloutions.
$\displaystyle x_1$	$=$	$\displaystyle 2$
$\displaystyle x_2$	$=$	$\displaystyle -2$

If you had not used the dashes, the solution would only have been $x = 2$ . So you would have lost the solution $x = - 2$ .

The Square Root

Examples

Square roots of terms

Examples

Domain of definition

Square roots and absolute values

How to handle roots of terms

Example 6x2\sqrt{6x^2}6x2​

Calculation rules

Making denominators rational

Taking square roots in equations

Übungsaufgaben: Square Roots

Example $\sqrt{6x^2}$