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Xarpolis
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Joined: 15 Oct 2002
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Location: Philly, PA
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 Posted: 08/06/05 - 11:00 Post subject: Algebra type question. I'm re-teaching myself it.
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Ok... There's a lot that I don't exactly remember about algebra, but here's what my question is.
little 4 ontop of the shelf for square root 58,000,000,000
How would you do that? going to attempt to drawl it out. remember... small 4
4\/58,000,000,000
Any ideas? the 4 should sit ontop of the \ in the square.
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sinrakin
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Joined: 11 Oct 2002
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Posted: 08/06/05 - 11:14 Post subject:
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Why the hell would you want to do roots manually? There is a longhand way to do it that we learned in grade school, which I think was this: http://mathforum.org/library/drmath/view/52610.html
There's also a way to do cube roots which is even harder, but I don't know if you can extend it to arbitrary roots - certainly not easily to fractional roots. The problem you stated you could coincidentally do by doint the square root procedure twice, since taking the square root twice is the same as the fourth root, but that was just lucky.
You can also use Newton's method, approximate or whatever. Or use a slide rule. Or use look it up in a table of logarithms, divide the logarithm by 4, and take the inverse log.
But I don't think doing roots by hand is going to teach you much about algebra really.
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Nictathan
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Posted: 08/06/05 - 11:14 Post subject:
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If I am not mistaken (and it's been a while since I did Alg as well) but that means that instead of it being the Sq root it's actually the ^4 root.
so say 16 in that same situation would be a 2 for the answer.
This could be completely wrong
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Xarpolis
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Joined: 15 Oct 2002
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Location: Philly, PA
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Posted: 08/06/05 - 11:17 Post subject:
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It's supposto be similar to a cube root, but with a 4 in place of the 3.
I'm just trying to figure this shit out.. pretty sure I slept through this part of high school.
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Brael
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Posted: 08/06/05 - 11:32 Post subject:
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87.26839695
If you have a graphing calculator it's 4 then the #sqrt key rather than just sqrt key. On a TI it's in a menu, math I think on a Casio (the better type) it's on the keypad probally next to the sqrt key.
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Xarpolis
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Posted: 08/06/05 - 11:41 Post subject:
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Ok... try explaining this one.
Simplify. Assume that x is positive
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Xarpolis
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Posted: 08/06/05 - 11:42 Post subject:
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and more importantly.. explain how you did it. This shit baffles me. it gives me 3 answers to choose from, and I can't logically figure my way to any of the 3
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Nictathan
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Joined: 11 Oct 2002
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Location: here... where I am... not with you
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Posted: 08/06/05 - 11:44 Post subject:
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Got me...
I need to go back to school and relearn alot of shit I think
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sinrakin
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Posted: 08/06/05 - 12:03 Post subject:
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K Xarp, first of all don't think of that V---- sign as the "square root" sign, it's the "root" sign. It means take n'th root of the number inside, where "n" is the number on the outside. It's just a convention that if you don't write a number "n" on the outside, you assume n=2, which means the square root. If the number you write the number 3, then it's the cube root.
An easier way to think of the n'th root of a number is that by definition it's the 1/n power of a number. So another way to write "the square root of 4" is 4 ^ 1/2 (where "^" means raise to the power). "the square root of 4" is the same as "4^.5" is the same as "2".
So in your gif, the first expression is "the square root of X to the fourth power". That's the same as "X ^(4 * 1/2)", which equals "X^2" or "X squared", since four times one half equals 2.
The second expression is "the cube root of X to the ninth" which means "X to the (9 times 1/3)" or "X to the third" since 9/3 = 3.
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Soriak
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Posted: 08/06/05 - 12:03 Post subject:
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square root of x^4 = x^2 and x^2 * x^2 = x^4
x^4 = x * x * x * x
and the 3rd root of x^9 = x^3 because x^3 * x^3 * x^3 = x^9
Or if you have numbers that mach up like here:
4 : 2 = 2
9 : 3 = 3
(divide the number above x with the nth root - n being 2 and 3 in these cases)
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Faerdal
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Posted: 08/06/05 - 12:08 Post subject:
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| Xarpolis wrote: | Ok... try explaining this one.
Simplify. Assume that x is positive
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you have two expressions multiplied together there... easiest thing is to break each one down as far as possible and combine like terms. this is going to be a pain in the ass with no root symbol but just follow my words.
the root of X^4 is just X^2
the third root of X^9 is (X^2 times third root of X)
combine the two X^2 to give X^4
final answer is X^4(times third root of X)
...anything outside the root symbol in a complex root can be handled as a normal algebraic expression and combined with like terms.
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Faerdal
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Posted: 08/06/05 - 12:09 Post subject:
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disregard me last.. it's been a while since i've done calculus and i appear to have gone retarded! soriak is indeed right, the third root of X^9 is X^3, not my answer =)
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Soriak
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Posted: 08/06/05 - 12:11 Post subject:
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ohh, that's one question...
I see it now =p
so you have square root of x^4 = x^2
and 3rd root of x^9 = x^3
multiply it:
x^2 * x^3 = x^5
if "assume that x is positive" didn't stand there, you'd have to keep in mind that the solution of any root with n being an even number (ie square root, 4th root etc) can also be negative.
ie the square root of x^2 could be x and -x
because:
x * x = x^2
-x * -x = x^2
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sinrakin
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Posted: 08/06/05 - 12:26 Post subject:
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Oh yeah I didn't notice that it was supposed to be one problem, not two. So the answer is x^2 * x^3 = x^5.
The funny thing is I don't see any reason why you should have to assume X is positive in this case. The sign of X doesn't change anything. But the funny thing I didn't think about is the answer is really "plus or minus x^5".
Since the answer of the left half is "plus or minus x^2". It doesn't depend on the sign of x; it just depends on the fact that taking the square root of x^4 yields two roots, x^2 and -(x^2).
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motherface
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Posted: 08/06/05 - 12:30 Post subject:
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The cube root if (x^9) is the same as (x^9)^(1/3). When raising an exponent to another exponent, you multiply the exponents. When multiplying exponents of the same base, you add the exponents
(x^3)^3 = (x^3) * (x^3) * (x^3) = (x^9)
(x^4)^(1/2) = (x^2)
(x^9)^(1/3) = (x^3)
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Soriak
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Posted: 08/06/05 - 12:31 Post subject:
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| sinrakin wrote: | | The funny thing is I don't see any reason why you should have to assume X is positive in this case. |
yes, it's necessary:
square root could be x^2 and -x^2
third root HAS to be +x^3 (because -x^3 * -x^3 * -x^3 is -x^9, not x^9)
so the results could be:
x^2 * x^3 = x^5
-x^2 * x^3 = -x^5
only even numbered roots have 2 solutions.
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Xarpolis
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Posted: 08/06/05 - 12:42 Post subject:
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Ok... Try this one now. Again: explain the steps as you do them. I'm like WAY off on this one, for some reason.
a=3, b=-8, c=5; find x
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Soriak
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Posted: 08/06/05 - 12:45 Post subject:
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ohh, as for the 4th root of 58,000,000,000, here's a little trick that comes in handy:
4th root of 58,000,000,000 is equal to the 4th root of 58 multiplied by the 4th root of 1,000,000,000
so, you could rewrite it as:
4V58 * 4V10^8
4V10^8 = 10^4
so
4V58 * 10^4
You can use sinrakin's way of figuring out the 4th root of 58, but on a question I'd leave it standing like that... easier to continue if there are further steps too.
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sinrakin
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Posted: 08/06/05 - 12:46 Post subject:
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| Soriak wrote: | | sinrakin wrote: | | The funny thing is I don't see any reason why you should have to assume X is positive in this case. |
yes, it's necessary:
square root could be x^2 and -x^2
third root HAS to be +x^3 (because -x^3 * -x^3 * -x^3 is -x^9, not x^9)
so the results could be:
x^2 * x^3 = x^5
-x^2 * x^3 = -x^5
only even numbered roots have 2 solutions. |
But making x positive doesn't change that.
If x=2, then you have sqrt(2^4) * cubert(2^9) = (+-4) * (8) = +-32.
If x=-2, then you have sqrt(-2^4) * cubert(-2^9) = (+-4) * (-8) = +-32.
The sign of x is irrelevent.
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motherface
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Posted: 08/06/05 - 12:47 Post subject:
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| Xarpolis wrote: | Ok... Try this one now. Again: explain the steps as you do them. I'm like WAY off on this one, for some reason.
a=3, b=-8, c=5; find x |
Uh, plug the numbers in and do the math?
Answer looks like 10/6 in the 10 seconds it took to do it in my head? The part that's being rooted evals to 4, so (8+2)/6.
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sinrakin
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Posted: 08/06/05 - 12:51 Post subject:
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| Xarpolis wrote: | Ok... Try this one now. Again: explain the steps as you do them. I'm like WAY off on this one, for some reason.
a=3, b=-8, c=5; find x |
That's the formula for one root of the quadratic equation, just to put it in context.
-b = 8
b^2 = 64
4ac = 60
2a = 6
so the answer is (8 + sqrt(64-60) ) / 6 = (8+sqrt(4))/6 = 10/6 = 1.666
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Soriak
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Posted: 08/06/05 - 12:52 Post subject:
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| Xarpolis wrote: | Ok... Try this one now. Again: explain the steps as you do them. I'm like WAY off on this one, for some reason.
a=3, b=-8, c=5; find x |
one step at a time:
-b + Vb * b - 4 * a * c
8 (becomes positive) + V8 * 8 - 4 * 3 * 5
8 + V64 - 60
8 + V4
8 + 2
10
2a = 2 * 3 = 6
x = 10/6
x = 5/3
edit: one more thing: you should leave 5/3 standing. Try to make it as small as possible (ie don't let 10/6 stand), but don't write down 1.6(period sign over 6)
Because if you have to continue with the term, it'll get messy and inaccurate.
Last edited by Soriak on 08/06/05 - 13:02; edited 1 time in total
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Xarpolis
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Posted: 08/06/05 - 13:01 Post subject:
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Sori's right. it was 5/3. I was getting the same thing Sin was, which is way off.
Hmm...
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sinrakin
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Posted: 08/06/05 - 13:07 Post subject:
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| Xarpolis wrote: | Sori's right. it was 5/3. I was getting the same thing Sin was, which is way off.
Hmm... |
Err what? Soriak and I (and Motherface) all got the same answer. Although given that this is actually an algebra course, Soriak is right, the literal answer they're looking for is probably 5/3, not 10/6 or 1.666 or whatever, although it sort of goes without saying, depending on what aspect of the answer you're trying emphasize.
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motherface
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Posted: 08/06/05 - 13:21 Post subject:
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Wow man, how long has it been since you took math?
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motherface
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Xarpolis
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Posted: 08/06/05 - 13:27 Post subject:
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5 years... I've been out of HS 5 years now, and finally going back to college for engineering. I just forgot a large chunk of algebra in the 5 years I havn't used it.
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motherface
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quotison
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Posted: 08/06/05 - 13:45 Post subject:
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| motherface wrote: | | Isn't there a special name for b^2 - 4ac? It's called "the ________". |
The discriminant, I believe.
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sinrakin
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Posted: 08/06/05 - 13:48 Post subject:
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But I thought discrimination was wrong?
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