Exercises: Terms with Square Roots

Determine the domain of definition

First determine the domain of definition.

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

Only positive expressions are allowed under the root, therefore only positive values or 0 may be used for $xx$.

$D={\mathbb{R}}_0^{+}D=\backslash mathbb\{R\}\_0^+$

Write the term without square roots

Conclude the roots with the calculation rules for square roots.

Determine the domain of definition

First determine the domain of definition.

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

Only positive expressions are allowed under the root.

The first root in the numerator is always positive because the variable $aa$ is squared.

However, in the second root of the numerator you may only insert positive values or 0 for $aa$.

The same applies to the root in the denominator, but here $aa$ must be greater than zero.

Therefore only positive values or 0 may be used for $aa$.

$D={\mathbb{R}}_{}^{+}D=\backslash mathbb\{R\}\_\{\}^+$, also $a>0a>0$

Write the term without square roots

Conclude the roots with the calculation rules for square roots.

Determine the domain of definition

First determine the domain of definition.

${\left(\sqrt{d-2}\right)}^2\backslash left(\backslash sqrt\{d-2\}\backslash right)^2$

Only positive expressions are allowed under the root. This gives you the following condition:

$d-2\ge 0d-2\backslash geq0$

$d\ge 2d\backslash geq2$

$D=[2;\mathrm{\infty })D=\backslash left[2;\backslash infty\backslash right)$

Write the term without square roots

Take the square root and do not forget the absolute value.

Determine the domain of definition

First determine the domain of definition.

$\sqrt{{(d-2)}^2}\backslash sqrt\{\backslash left(\; d-2\; \backslash right)^2\}$

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 $dd$.

$D=\mathbb{R}D=\backslash mathbb\{R\}$

Write the term without square roots

Only positive or zero expressions are allowed under the root. So $x-36x-36$ must be positive or zero. $x-36x-36$ becomes zero for $x=36x\; =\; 36$.

• When we make $xx$ smaller (so ), then $x-36x-36$ gets negative (which is not good)

• When we make $xx$ larger (so $x>36x>36$), then $x-36x-36$ becomes positive (which is what we want).

So $x\ge 36x\; \backslash ge\; 36$ is the required condition for obtaining a positive or zero number under the root. Further, $xx$ is a real number, so

Only positive or zero expressions are allowed under the root. So $36+x^{2}36+x^2$ must not be negative. Now, $x^{2}x^2$ is never negative and if we add $+36+36$, it only becomes "more positive". So any $xx$ is allowed in the domain of definition, as long as it is a real number:

Only positive or zero expressions are allowed under the root. So $x+36x+36$ must not be negative. Further, we cannot divide by zero, which excludes the case $\sqrt{x+36}=0\backslash sqrt\{x+36\}\; =\; 0$, which happens if $x+36x+36$ becomes zero. So only $x+36>0x+36\; >\; 0$ is allowed. This is the case whenever $x>-36x>-36$. Further, we know that $xx$ is a real number.

So the domain of definition of $\frac{1}{\sqrt{\mathrm{x}+36}}\backslash frac1\{\backslash sqrt\{\backslash mathrm\; x+36\}\}$ is characterized by

Only positive or zero expressions are allowed under the root. So $x^2-36x^2-36$ must not be negative. That means, $x^{2}x^2$ must be larger or equal $3636$.

• For $x=6x\; =\; 6$ and $x=-6x\; =\; -6$, you exactly obtain $x^2=36x^2\; =\; 36$. This is a "good case"

• If $xx$ is positive, then making $xx$ larger (so $x>6x>6$) will make $x^{2}x^2$ larger than $3636$, which is also a "good case".

• Analogously, for negative $xx$, making $xx$ even more negative (so ) will also result in $x^2>36x^2\; >\; 36$, which is good for our purposes.

• However, if -, so $x\in (-6,6)x\; \backslash in\; (-6,\; 6)$, then $x^{2}x^2$ becomes smaller than 36, which is not good.. So we have to exclude this case.

Further, we know that $xx$ is a real number. So the domain of definition is given by

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

This expression is not defined whenever there is a negative number under the root.

Both exponents of $aa$ and $bb$ are even, so $a^{4}a^4$ and $b^{2}b^2$ are always positive or zero. Thus, the expression under the root is also positive or zero and the root is always defined:

$D_{\max }=\mathrm{R}D\_\{\backslash max\}=ℝ$

Now, take the root.

$\sqrt{49a^4{b}^2}=7\cdot a^2\cdot \mid b\mid \backslash sqrt\{49a^4b^2\}=7\backslash cdot\; a^2\backslash cdot\backslash left|b\backslash right|$

You have to put dashes, since $bb$ could be negative.

$\sqrt{{(-b)}^2}\backslash sqrt\{\backslash left(-b\backslash right)^2\}$

The root is net defined, if the expression under the root is negative. Now, squares of any number (also of $-b-b$) are always positive of zero, so there are never any problems with the definition of the root.

$D_{\max }=\mathrm{R}D\_\{\backslash max\}=ℝ$

Square $-b-b$ .

${\sqrt{-b}}^2\backslash sqrt\{-b\}^\{\backslash ;2\}$

The number under the root must not be negative. $\Rightarrow \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;$ Only negative numbers or 0 are allowed for $bb$.

$D_{\max }={\mathrm{R}}_0^{-}D\_\{\backslash max\}=ℝ\_0^-$

${\sqrt{-b}}^2\backslash sqrt\{-b\}^\{\backslash ;2\}$$=-b=-b$

So finally, power and root cancel.

$\sqrt{{(1-2x)}^2}\backslash sqrt\{\backslash left(1-2x\backslash right)^2\}$

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.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{{(1-2x)}^2}=\mid 1-2x\mid \backslash sqrt\{\backslash left(1-2x\backslash right)^2\}=\backslash left|1-2x\backslash right|$

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.

$\sqrt{{(x-y)}^2}\backslash sqrt\{\backslash left(x-y\backslash right)^2\}$

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.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{{(x-y)}^2}=\mid x-y\mid \backslash sqrt\{\backslash left(x-y\backslash right)^2\}=\backslash left|x-y\backslash right|$

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.

$\sqrt{x^2+y^2}\backslash sqrt\{x^2+y^2\}$

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.

${\mathrm{D}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathbb{R}\{\backslash mathrm\; D\}\_\backslash mathrm\{max\}=\backslash mathbb\{R\}$

$\sqrt{x^2+y^2}=\sqrt{x^2+y^2}\backslash sqrt\{x^2+y^2\}=\backslash sqrt\{x^2+y^2\}$

It is not possible to further simplify this term. It is already in the simplest possible form.

$\sqrt{x^2\cdot y^2}\backslash sqrt\{x^2\backslash cdot\; y^2\}$

Both $x^{2}x^2$ and $y^{2}y^2$ are never zero. The same holds for the product, so there are never any problems with definition of the square root.

$\to \mathbb{D}=\mathbb{R}\backslash rightarrow\; \backslash mathbb\{D\}=\; \backslash mathbb\{R\}$

Now, $\sqrt{x^2\cdot y^2}=\sqrt{{(x\cdot y)}^2}\backslash sqrt\{x^2\backslash cdot\; y^2\}=\backslash sqrt\{\backslash left(x\backslash cdot\; y\backslash right)^2\}$. Take the root. Root and square cancel each other out, but it is necessary to retain absolute values, since $xx$ or $yy$ might be negative and turned positive by the squaring.

$=\mid x\cdot y\mid =\backslash left|x\backslash cdot\; y\backslash right|$

For which $xx$ do we have $\sqrt{x^2}=-x\backslash sqrt\{x^2\}=-x$?

In general, we have $\sqrt{x^2}=\mid x\mid \backslash sqrt\{x^2\}=\backslash left|x\backslash right|$. So:

For which $xx$ is it true that $\mid x\mid =-x\backslash left|x\backslash right|=-x$?

Answer: For all values of $xx$ that are less than or equal to $00$.

This is true for all values $x\ge 1x\backslash geq1$ , since $\sqrt{{(x-1)}^2}=\mid x-1\mid \backslash sqrt\{\backslash left(x-1\backslash right)^2\}=\backslash left|x-1\backslash right|$ and the absolute value function is equal to the right-hand side exactly for $x-1\ge 0x-1\backslash geq0$ .