Dividing Radical Expressions

Dividing Radical Expressions exercise

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

  Hints

  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.

  Solution

  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.

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

  Hints

  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: $\sqrt{a^2}=a$.

  Solution

  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. $\frac{-6\sqrt{20}}{2\sqrt5}=-3\times\sqrt{4}=-3\times 2=-6$

• Find the length of the hypotenuse $c$.

  Hints

  $c$ 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: $\sqrt{a^2}=a$.

  Solution

  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.

  Hints

  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: $\sqrt{a^2}=a$.

  Solution

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

  $~$

  Let's start with the equation in the upper left-hand corner:

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

  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.

  Let's start with the equation in the upper right-hand corner:

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

  The last equation is the one in the bottom right-hand corner:

  $\begin{array}{rclll} \frac{-2\sqrt{48}}{5\sqrt3}&=&~\frac{-2}{5}\times\frac{\sqrt{48}}{\sqrt{3}}&|&\text{seperating the coefficients}\\ &=&~-\frac25\times\sqrt{\frac{48}{3}}&|&\text{quotient property}\\ &=&~-\frac25\times\sqrt{16}\\ &=&~-\frac25\times4=-\frac85&|&\text{as }4^2=16 \end{array}$

• Simplify the given radical expressions.

  Hints

  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:

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

  Solution

  Here we study the quotient property of square roots,

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

  or rather

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

  Let's start with

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

  Here you see another example:

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

• Identify the simplified form of the radical expressions.

  Hints

  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: $\sqrt{a^2}=a$.

  Solution

  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:

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

  $~$

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