# Thread: Formulas and Inequalities (Problem Solving)

1. ## Formulas and Inequalities (Problem Solving)

Hi All,
In the following question

The competitive edge of a baseball team is defined as $\displaystyle \sqrt{\frac{W}{L}}$. W = number of wins, L = number of losses. This year a the team had 3 times as many wins and $\displaystyle \frac{1}{2}$ as many losses. By what factor did the competitive edge increase?

1. $\displaystyle c = \sqrt{\frac{W}{L}}$ where c = competitive edge.
2. Need to determine the original competitive edge so pick numbers $\displaystyle c = \sqrt{\frac{2}{4}}$
3. Simplify $\displaystyle c = \sqrt{\frac{2}{4}}$ = $\displaystyle c = \sqrt{\frac{1}{2}}$ = $\displaystyle c = \frac{\sqrt{1}}{\sqrt{2}}$ = $\displaystyle c = \frac{1}{\sqrt{2}}$

QUESTION ONE. I would typically stop the symplification here. However the question went on to state $\displaystyle c = \frac{\sqrt{2}}{2}$ I understand HOW this next step was derived. What I'm trying to understand is WHY (i.e how did the author know the process required this extra step)?

4. $\displaystyle W.3$ $\displaystyle L.\frac{1}{2}$ = $\displaystyle c = \sqrt{\frac{6}{2}} = \sqrt{3}}$

5. $\displaystyle \frac{\sqrt{2}}{2}X = \sqrt{3}}$ = $\displaystyle \frac{\sqrt{3}}{1}.\frac{2}{\sqrt{2}} = \frac{2_\sqrt{3}}{\sqrt{2}}$

QUESTION TWO. I would stop sympifying here. However the question went on to state $\displaystyle \frac{2_\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}}$ $\displaystyle X = \sqrt{3}\sqrt{2}$. Again I understand the HOW. What I'm tring to understand is WHY (i.e. how did the author know to simplify it further)?

I hope that makes sense.

Thanks,
D

2. Originally Posted by dumluck
Hi All,
In the following question

The competitive edge of a baseball team is defined as $\displaystyle \sqrt{\frac{W}{L}}$. W = number of wins, L = number of losses. This year a the team had 3 times as many wins and $\displaystyle \frac{1}{2}$ as many losses. By what factor did the competitive edge increase?

1. $\displaystyle c = \sqrt{\frac{W}{L}}$ where c = competitive edge.
2. Need to determine the original competitive edge so pick numbers $\displaystyle c = \sqrt{\frac{2}{4}}$
3. Simplify $\displaystyle c = \sqrt{\frac{2}{4}}$ = $\displaystyle c = \sqrt{\frac{1}{2}}$ = $\displaystyle c = \frac{\sqrt{1}}{\sqrt{2}}$ = $\displaystyle c = \frac{1}{\sqrt{2}}$

QUESTION ONE. I would typically stop the symplification here. However the question went on to state $\displaystyle c = \frac{\sqrt{2}}{2}$ I understand HOW this next step was derived. What I'm trying to understand is WHY (i.e how did the author know the process required this extra step)?
I doubt that any one is saying that "rationalizing the denominator" is required. But it is true that in working with fractions (especially adding fractions where you have to get "common denominators") it is best to have the denominator of the fraction as simple as possible.

4. $\displaystyle W.3$ $\displaystyle L.\frac{1}{2}$ = $\displaystyle c = \sqrt{\frac{6}{2}} = \sqrt{3}}$

5. $\displaystyle \frac{\sqrt{2}}{2}X = \sqrt{3}}$ = $\displaystyle \frac{\sqrt{3}}{1}.\frac{2}{\sqrt{2}} = \frac{2_\sqrt{3}}{\sqrt{2}}$

QUESTION TWO. I would stop sympifying here. However the question went on to state $\displaystyle \frac{2_\sqrt{3}\sqrt{2}}{\sqrt{2}\sqrt{2}}$ $\displaystyle X = \sqrt{3}\sqrt{2}$. Again I understand the HOW. What I'm tring to understand is WHY (i.e. how did the author know to simplify it further)?

I hope that makes sense.

Thanks,
D
I presume that the author immediately saw the $\displaystyle \sqrt{2}$ in both numerator and denominator and, of course canceled them. Would you prefer "$\displaystyle \frac{3}{3}$" or "1"? Also, since $\displaystyle 2= \sqrt{2}\sqrt{2}$, $\displaystyle \frac{2}{\sqrt{2}}= \frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}= \sqrt{2}$ which is a much simpler form.