# Where to get algebraic help?June 17, 2014 8:01 PM   Subscribe

I am working through the mooculus calculus one mooc as a refresher (which I clearly need), and there is one thing I have run across that I just don't see. Can you help me figure out what is going on, and what I can do to work out future problems?

In the video on Inverses of Functions, at about 7:00, the professor simplifies:

(6 +/- sqrt(6^2 - 4(-12 + h/5))/2

to

3 +/- sqrt(9 + 12 -h/5)

and I just can't see how that is done. What is confusing me is the term "+12 - h/5" in the simplified square root. It seems to be off by a factor of two, since the -4/2 = -2, not -1. What am I missing in this case?

In addition, I am going to be working through this over the next weeks and have no doubt I will run into similar problems. Is there a website or software that will perform this type of simplification and show the steps? I have access to Matlab and Mathematica, but have never used them, if they help.
posted by carpographer to Science & Nature (5 answers total) 2 users marked this as a favorite

The 2 in the denominator is out of the square root. Thus when you take it into the sqrt you get 4 not 2.

To be more explicit:

(6 +/- sqrt(6^2 - 4(-12 + h/5))/2
=(3 +/- sqrt(36/4 - 4/4(-12 + h/5)))
=3 +/- sqrt(9 + 12 -h/5)
posted by peacheater at 8:06 PM on June 17, 2014 [1 favorite]

6^2 - 4(-12 + h/5) =

36 +48 +4h/5

Divide it through by 2^2, basically, since it's inside a square root.
posted by empath at 8:08 PM on June 17, 2014

I realize that I may not have made it super clear above what I mean by taking something into a sqrt. Imagine that the numerator is anything within a square root. For example:
sqrt(6)/2

This can be rewritten as sqrt(6)/sqrt(4) (since 2^2 is 4). Since both the numerator and denominator are under sqrt, we can rewrite this as sqrt(6/4) = sqrt(1.5).

It's exactly the same logic that operates above, it's just a bit more complicated numerically.
posted by peacheater at 8:09 PM on June 17, 2014 [2 favorites]

you can try looking on Khan Academy to see if they have videos that are relevant for you. They have a ton of math and science videos as supplements to a topic, and they're usually short and easy to understand. worth a look!
posted by carlypennylane at 8:54 PM on June 17, 2014 [2 favorites]

Seconding the bit about Khan Academy, but also adding that the site also has a section where you can do exercises and get immediate feedback on whether or not you got them correct. The videos are definitely worth a look too and are generally clearer most of my math teachers/professors in high school and college.
posted by bluekazoo at 12:53 AM on June 18, 2014

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