Sunday, 4 May 2014

Weird square root disappearing and flipping fraction upside down?



So here I was, making 2 math problems, I was able to solve them, but 2 operations seem a bit intractable to me. Maybe you can help me understand why this is true:



The first problem:
$$x = \frac{1}{5} - \frac{4}{y}$$

$$\frac{4}{y} =\frac{1}{5} - x$$



$$\frac{4}{y} = \frac{1-5x}{5} $$



$$\frac{y}{4} = \frac{5}{1-5x}$$



Why is it possible to turn $\frac{y}{4}$ upside down?



$$y = 20 / 1 - 5 x$$




The second problem:



$$4A√B - √B = 3$$



$$√B(4A-1) = 3$$



Where does the -√B go? I understand that the -1 comes from the - sign in front of the square root. But where does the other √B go?



$$\sqrt{B} = \frac{3}{(4A-1)}$$




$$B = (\frac{3}{(4A-1)} )^2$$



$$B = \frac{9}{(4A-1)^2}$$



Everything, except above the bold text I understand.
Maybe I do not understand the full extent of a certain rule which I am familiar with in simpler situations. That's why I think an example would be very useful. I really want to have a deep understand of why these things are true.



Greetings,
Bowser.


Answer




First question: if two fractions are equal, then their reciprocals are equal too (the reciprocal of a fraction is the same fraction "turned upside down", using your terminology).



Second question: the $B$ is still there: if you multiply $\sqrt{B}(4A-1)$ you get indeed $4A\sqrt{B}-\sqrt{B}$. It's called "distributive law".


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...