Really need help understanding simplifying a promlem

[uk2la99]

Hi all,

I have been brushing up on my Trigonometry and whilst taking some practice questions on a website, I couldn't seem to get the right answer to the final simplification of an equation : 5/ 5√61/61 ... according to the test help this simplifies to √61. Can someone please explain how on earth this happens in detail? I understand the 5s canceling out and that would leave √61/61 in the denominator. But then ??? I haven't a clue : ( Please help and fully explain. Thanks so much.

 

[JeffM]

Hi all,

I have been brushing up on my Trigonometry and whilst taking some practice questions on a website, I couldn't seem to get the right answer to the final simplification of an equation : 5/ 5√61/61 ... according to the test help this simplifies to √61. Can someone please explain how on earth this happens in detail? I understand the 5s canceling out and that would leave √61/61 in the denominator. But then ??? I haven't a clue : ( Please help and fully explain. Thanks so much.

EDIT: THIS ANSWER IS ALL SCREWED UP. PAY NO ATTENTION WHATSOEVER TO THE INCORRECT CALCULATIONS BELOW


\(\displaystyle \dfrac{\dfrac{5}{5\sqrt{61}}}{61} = \dfrac{5}{5 \sqrt{61}} * \dfrac{61}{1} = \dfrac{61}{\sqrt{61}} =\dfrac{61}{\sqrt{61}} * \dfrac{\sqrt{61}}{\sqrt{61}} = \dfrac{61 * \sqrt{61}}{\sqrt{61} * \sqrt{61}} = \dfrac{61 * \sqrt{61}}{61} = \sqrt{61}. \)

It's called rationalizing the denominator.

Edit: HallsofIvy is right about the parentheses. You DO need them to present your problem correctly.

 

[HallsofIvy]

I would disagree with JeffM- the parentheses are "technically necessary"!

He interpreted your "5/ 5√61/61" as 5/ (5√61/61),\(\displaystyle \frac{5}{\left(\frac{5\sqrt{61}}{61}\right)}\), perhaps because of that space after the first "/".
Dividing by a fraction is the same as multiplying by its reciprocal: \(\displaystyle 5\left(\frac{61}{5\sqrt{61}}\right)= \sqrt{61}\)
The "5"s cancel and, because \(\displaystyle 61= (\sqrt{61})^2= \sqrt{61}\sqrt{61}\), \(\displaystyle \frac{61}{\sqrt{61}}= \frac{\sqrt{61}\sqrt{61}}{\sqrt{61}}= \sqrt{61}\)

But it could be just as easily be interpreted as \(\displaystyle \frac{\left(\frac{5}{5\sqrt{61}}\right)}{61}\). Since now the denominator is not a fraction, this the same as \(\displaystyle \frac{5}{5\sqrt{61}}\frac{1}{61}= \frac{1}{61\sqrt{61}}= \frac{1}{\sqrt{61^3}}\).

Of course, since you say "this simplifies to √61" we know it must be the first.

 

[uk2la99]

Some parentheses would be nice, not because they are technically necessary but to ensure that we read the problem correctly.

\(\displaystyle \dfrac{\dfrac{5}{5\sqrt{61}}}{61} = \dfrac{5}{5 \sqrt{61}} * \dfrac{61}{1} = \dfrac{61}{\sqrt{61}} =\dfrac{61}{\sqrt{61}} * \dfrac{\sqrt{61}}{\sqrt{61}} = \dfrac{61 * \sqrt{61}}{\sqrt{61} * \sqrt{61}} = \dfrac{61 * \sqrt{61}}{61} = \sqrt{61}. \)

It's called rationalizing the denominator.

Edit: HallsofIvy is right about the parentheses. You DO need them to present your problem correctly.

Wow... thanks so much, very clear and I can see I just didn't work it through enough.