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Author Topic: Math Help. (Again!)  (Read 609 times)

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Offline WolfyTopic starter

Math Help. (Again!)
« on: November 02, 2010, 09:45:58 PM »
I needs it. o-o

specifically, for problems like:

y^2= x+2


(Square root= 4x-9)=1

(Cube root= 3x-1)=-4

D: any help would be appreciated.

Offline Trieste

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Re: Math Help. (Again!)
« Reply #1 on: November 03, 2010, 08:33:26 AM »
I assume you're needing values for x and y?

Online Oniya

Re: Math Help. (Again!)
« Reply #2 on: November 03, 2010, 08:42:24 AM »
I'm assuming you're solving for x in these?

The first one is rather troublesome if you are, since you have an x in the exponents.

The second one is very straight-forward.  Start out by swapping the two sides (so x is on the left side), and then isolate x as normal.
--If you are solving this equation for y, it's important to remember that every positive number has two square roots, one positive and one negative, and that taking the square root of both sides applies to the whole side, so you'd end up with two answers.  One would look like the third example, and the other would be the negative of that.

With the third and fourth, you don't have to worry about that, because while every positive number has two square roots, there's only one answer you can get by squaring (or cubing) a given number.  Square (or cube) both sides to get rid of the root symbol, then isolate x.

Offline WhiteyChan

Re: Math Help. (Again!)
« Reply #3 on: November 04, 2010, 12:01:24 PM »
Wolfram-Alpha is your friend :D

In all seriousness, the first one is pretty tricky, as Oniya said. I'm not sure you can get a single equation for y in terms of x from that. If you can, it'll most likely involve logs to get rid of the power of x term.

Is the second one (y2 = x + 2) a simultaneous equation with the first, or a separate one? If its a separate equation, its easy as - take the square root of both sides to get y in terms of x, or just minus 2 from both sides to get y in term of x. If its a simultaneous equation with the first... Its still not really doable.

The third and fourth ones are easy, as Oniya said as well. Square/cube both sides and rearrange.

Don't hesitate to send me a PM if you need more help - I'm doing a degree in physics, I'm pretty sure I can help with anything you need.

Offline minerva7790

Re: Math Help. (Again!)
« Reply #4 on: November 05, 2010, 10:03:03 AM »
Wait... They're not simultaneous equations, are they? The first pair of equations?