Video Transcript

Meet the man they call Wendelin
the path he takes is bafflin'.
The loyal, royal courier is he,
With no time to wander,
his travels take him yonder,
trundlin' packages for his Lord and Lady.
For a full day,
he travels the way
from Arden to Barton via Circuity.
Taking the long way 'round,
there's sure to be found,
a way shorter than his route currently.
What do I spy?
Is there a new bridge nearby?
Is it really a shortcut, Wendelin wonders.
To figure this out,
and leave no doubt,

Rather than traveling two-thirds of a day to Circuity, to trip trap 'cross the bridge, and then travel an additional one-third of a day from there to Barton, Wendelin can simply cross the new bridge and travel directly from Arden to Barton! How's that, you ask?

### The Pythagorean Theorem

Remember the Pythagorean Theorem? ‘a’ squared plus ‘b’ squared is equal to ‘c’ squared. Side 'c' is called the hypotenuse and is the longest side! Using the Pythagorean Theorem to solve for the missing measurement, Wendelin knows how to set up the equation to solve for a positive rational expression, but he needs help simplifying the radical expression.

### The Quotient Property of Square Roots

To solve this type of problem, use the Quotient Property of Square Roots: The square root of a quotient is equal to the quotient of the square roots of the numerator and denominator. For all positive real numbers 'a' and 'b' - 'b' does not equal zero. Take a look at how we can apply the quotient property of square roots to solve this problem: The square root of 90 divided by the square root of 10. This is the same as the square root of the fraction 90 over 10. We can simplify this to the square root of 9, and then solve. The quotient is 3, since squaring this number gives you 9.

Let’s try another problem: the square root of the fraction 49 over 9. Hmm, simplifying the fraction won't help, but we can split up the parts and take the square roots. The square root of 49 is equal to 7, and the square root of 9 is equal to 3 giving us 7 over 3. For more complex fractions such as this, note how we separate the coefficients from the radicals, and then apply the quotient property to simplify the square root. The fraction under this radical is simplified to the square root of 4, and the rest is easy.

Getting back to the original problem. Wendelin uses what he’s learned to simplify the radical expression. Using the Quotient Property of Square Roots, the solution is the square root of 5 over 3. Compared to the old route, the new route will save him one-quarter of a day!

Wendelin makes his merry way,
along his newfound pathway
to try out the new overpass.
He notices a troll,
who's blocking his goal,
with a lot of zeal and a lot of sass.
Wendelin solves the problems effortlessly,
but tomorrow's another story.

### Du möchtest dein gelerntes Wissen anwenden? Mit den Aufgaben zum Video Dividing Radical Expressions kannst du es wiederholen und üben.

• #### Calculate the time it takes to travel the dotted path.

##### Tipps

The Pythagorean Theorem states that

$a^2+b^2=c^2$,

where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the legs.

The quotient property of square roots:

To use the Pythagorean theorem first consider the hypotenuse and the legs.

##### Lösung

The given situation shows a right triangle with the two legs of lengths $\frac23$ and $\frac13$. The length of the hypotenuse $c$ is what we want to find.

We can use the Pythagorean Theorem to find $c$:

$a^2+b^2=c^2$

with

• $a=\frac13$
• $b=\frac23$
So let's put the known values in:

$c^2=\left(\frac13\right)^2+\left(\frac23\right)^2$.

Taking the square root, we get

$c=\sqrt{\left(\frac13\right)^2+\left(\frac23\right)^2}$.

We then have that

$\begin{array}{rcl} c&=&~\sqrt{\left(\frac13\right)^2+\left(\frac23\right)^2}\\ &=&~\sqrt{\frac19+\frac49}\\ &=&~\sqrt{\frac59} \end{array}$

And now we use the quotient property of square roots,

$\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}$,

to get

$c=\frac{\sqrt5}{\sqrt9}=\frac{\sqrt5}{3}\approx 0.75$.

Thus we find that the new path is $0.75$ as long as the older one. Or, the other way round, Wendelin saves $0.25$ of a day.

• #### Simplify the given radical expressions.

##### Tipps

The square root of any square is the squared term itself:

• $\sqrt4=\sqrt{2^2}=2$
• $\sqrt{16}=\sqrt{4^2}=4$
• $\sqrt{25}=\sqrt{5^2}=5$

You can simplify any fraction by dividing the numerator as well as the denominator by the same value:

$\frac{16}{8}=\frac{16\div8}{8\div8}=\frac21=2$

##### Lösung

Here we study the quotient property of square roots,

$\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}$,

or rather

$\frac{\sqrt a}{\sqrt b}=\sqrt{\frac ab}$.

$\begin{array}{rclll} \frac{\sqrt{90}}{\sqrt{10}}&=&~\sqrt{\frac{90}{10}}&|&\text{ quotient property}\\ &=&~\sqrt 9&|&\text{ simplifying the fraction}\\ &=&~3&|&\text{ because }3^2=9 \end{array}$

Here you see another example:

$\begin{array}{rclll} \sqrt{\frac{49}{9}}&=&~\frac{\sqrt{49}}{\sqrt 9}&|&\text{ quotient property}\\ &=&~\frac73&|&\text{ because }7^2=49 \text{ and }3^2=9 \end{array}$

• #### Explain how to simplify $(-6\sqrt{20})\div(2\sqrt 5)$.

##### Tipps

Keep in mind that $\frac{a\times b}{c\times d}=\frac{a}{c}\times\frac{b}{d}$.

The quotient property of square roots:

The square root of any square is the squared term itself: $\sqrt{a^2}=a$.

##### Lösung

We can simplify this complicated looking fraction in a few steps:

1. First we separate the coefficients from the radicals.
2. Next apply the quotient property of square roots to simplify the square roots.
3. Lastly simplify all terms as much as possible.
Let's try it out:
1. $\frac{-6\sqrt{20}}{2\sqrt5}=\frac{-6}2\times\frac{\sqrt{20}}{\sqrt5}$
2. $\frac{-6\sqrt{20}}{2\sqrt5}=-3\times\sqrt{\frac{20}5}$
3. $\frac{-6\sqrt{20}}{2\sqrt5}=-3\times\sqrt{4}=-3\times 2=-6$

• #### Identify the simplified form of the radical expressions.

##### Tipps

To seperate the coefficients, if necessary, use $\frac{a\times b}{c\times d}=\frac{a}{c}\times\frac{b}{d}$.

Here you see an example for the quotient property of square roots:

The square root of any square is the squared term itself: $\sqrt{a^2}=a$.

##### Lösung

If you have to divide radical expressions or take the square root of fractions you have to use the quotient property of square roots: $\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}$.

$~$

Once again we practice this, together with separating the coefficients first, if necessary:

$\begin{array}{rclll} \frac{\sqrt{242}}{\sqrt2}&=&~\sqrt{\frac{242}{2}}&|&\text{quotient property}\\ &=&~\sqrt{121}\\ &=&~11&|&\text{because }11^2=121 \end{array}$

$~$

$\begin{array}{rclll} \sqrt{\frac{121}{25}}&=&~\frac{\sqrt{121}}{\sqrt{25}}&|&\text{quotient property}\\ &=&~\frac{11}{5}&|&\text{because }11^2=121\text{ and }5^2=25 \end{array}$

$~$

For the next examples we first have to separate the coefficients:

$\begin{array}{rclll} \frac{12\sqrt{80}}{3\sqrt5}&=&~\frac{12}3\times\frac{\sqrt{80}}{\sqrt{5}}&|&\text{seperating the coefficients}\\ &=&~4\times\sqrt{\frac{80}{5}}&|&\text{quotient property}\\ &=&~4\times\sqrt{16}\\ &=&~4\times4=16&|&\text{because }4^2=16 \end{array}$

$~$

$\begin{array}{rclll} \frac{5\sqrt{112}}{20\sqrt7}&=&~\frac{5}{20}\times\frac{\sqrt{112}}{\sqrt{7}}&|&\text{seperating the coefficients}\\ &=&~\frac14\times\sqrt{\frac{112}{7}}&|&\text{quotient property}\\ &=&~\frac14\times\sqrt{16}\\ &=&~\frac14\times4=1&|&\text{because }4^2=16 \end{array}$

• #### Find the length of the hypotenuse $c$.

##### Tipps

$c$ is the length of the hypotenuse. The lengths of the legs are given.

Here you see the Pythagorean theorem:

$a^2+b^2=c^2$.

Take the square root last.

The square root of any square is the squared term itself: $\sqrt{a^2}=a$.

##### Lösung

The tool we need is the Pythagorean theorem, which states that

$a^2+b^2=c^2$,

where $a$ and $b$ are the lengths of the legs of a right triangle and $c$ is the hypotenuse.

In all the examples above the hypotenuse is wanted.

We are asked the hypotenuse of several right triangles, where the lengths of the legs are known:

• $a=4$ and $b=3$: So we get $c=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
• $a=7$ und $b=4$ give us $c=\sqrt{7^2+4^2}=\sqrt{49+16}=\sqrt{65}$
• $a=5$ and $b=\sqrt{11}$ lead to $c=\sqrt{5^2+\sqrt{11}^2}=\sqrt{25+11}=\sqrt{36}=6$
• $a=\frac14$ and $b=\frac13$ result in $c=\sqrt{\left(\frac35\right)^2+\left(\frac45\right)^2}=\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{\frac{25}{25}}=1$

• #### Solve the radical equations.

##### Tipps

Use the quotient property of square roots each time.

For the two equations on the right you first have to seperate the coefficients from the radicals.

The square root of any square is the squared term itself: $\sqrt{a^2}=a$.

##### Lösung

For the following equations we use the quotient property of square roots: $\sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b}$.

$~$

$\begin{array}{rclll} \frac{\sqrt{24}}{\sqrt6}&=&~\sqrt{\frac{24}{6}}&|&\text{quotient property}\\ &=&~\sqrt4\\ &=&~2&|&\text{as }2^2=4 \end{array}$

Now for the equation in the bottom left-hand corner:

$\begin{array}{rclll} \sqrt{\frac{49}{100}}&=&~\frac{\sqrt{49}}{\sqrt{100}}&|&\text{quotient property}\\ &=&~\frac7{10}&|&\text{as }7^2=49\text{ and }10^2=100 \end{array}$

$~$

To solve the equations on the right side we first have to separate the coefficients.

$\begin{array}{rclll} \frac{3\sqrt{24}}{12\sqrt6}&=&~\frac3{12}\times\frac{\sqrt{24}}{\sqrt{6}}&|&\text{seperating the coefficients}\\ &=&~\frac14\times\sqrt{\frac{24}{6}}&|&\text{quotient property}\\ &=&~\frac14\times\sqrt4\\ &=&~\frac14\times2=\frac12&|&\text{as }2^2=4 \end{array}$
$\begin{array}{rclll} \frac{-2\sqrt{48}}{5\sqrt3}&=&~\frac{-2}{5}\times\frac{\sqrt{48}}{\sqrt{3}}&|&\text{seperating the coefficients}\\ &=&~-\frac25\times\sqrt{\frac{48}{3}}&|&\text{quotient property}\\ &=&~-\frac25\times\sqrt{16}\\ &=&~-\frac25\times4=-\frac85&|&\text{as }4^2=16 \end{array}$