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I got X = sqrt (3ay + z^2)

I'm pretty sure you can't break down the z^2 because the square root must go over the whole term 3ay + z^2. Am I right?

2006-11-07 00:40:14 · 4 answers · asked by Allen S 2 in Science & Mathematics Mathematics

4 answers

yes, you are correct.

2006-11-07 00:44:32 · answer #1 · answered by JEREMY 2 · 0 0

Unfortunately, your question isn't clear. :( What is it you're looking for? What are you supposed to solve for or determine?

If you're solving for y, then the answer obviously is:
y = (x^2 – z^2)/3a

If you're solving for x, then the answer, as you pointed out, is:
x = \|3ay + z^2

And yes, you are right, the square root operation gets performed on the whole term. In a situation, if you're not sure about this, an easy way to know what to do is to remember that what you're really doing is applying the square root operation to everything on either side of the equation. For example:

For the equation: w^2 + x^2 = y^2 + z^2, if you square root both sides, you get: \|w^2 + x^2 = \|y^2 + z^2.

Since you're square rooting both sides, then regardless how many terms there are on either side, they're considered to be one group, since that group consititutes either side of the equation. i.e.: the right side of my equation is y^2 + z^2, not y^2 or z^2.


If you're solving for z, then the answer is thus:
z = \|x^2 – 3ay

I am quite surprised by your question. There is only 1 equation, 4 unknowns and the desired aim is equally unknown! A most unusual situation! I sure hope that helped. :) Take care and have a great day. :)

2006-11-07 00:57:11 · answer #2 · answered by Cogano 3 · 0 0

you are absolutely right

2006-11-07 00:47:40 · answer #3 · answered by grandpa 4 · 0 0

you are absolutely right

2006-11-07 00:45:00 · answer #4 · answered by raj 7 · 0 0

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