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#1 2005-05-18 07:43:53
- im really bored
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Last one hopefully
Find the square root of (3/2)(x-1) + (sqrt(2x^2 - 7x - 4))
its not equal to anything so its all about manipulation, thats about all I can help lol.
#3 2005-05-18 07:58:46
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Re: Last one hopefully
This is as far as I can get:
(3/2)(x-1) + sqrt(2x^2 - 7x - 4))
(3/2)(x-1) + sqrt(2x^2 - 8x + x - 4))
(3/2)(x-1) + sqrt(2x(x-4) + x - 4))
(3/2)(x-1) + sqrt((2x+1)(x-4))
(3/2)(x-1) + sqrt(2x+1)*sqrt(x-4) (best I can do)
Try squaring:
[(3/2)(x-1)]^2 + 2 * (3/2)(x-1) * sqrt(2x+1)*sqrt(x-4) + (2x+1)*(x-4)
9/4 * (x-1)^2 + 3 * (x-1) * sqrt(2x+1)*sqrt(x-4) + (2x+1)*(x-4)
Nope, didn't do any good. Ah well ....
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#4 2005-05-18 09:39:42
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Re: Last one hopefully
Im not having any better luck with this problem
#5 2005-05-18 18:38:28
- scobbysteve
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Re: Last one hopefully
i hate maths if hitler was maths, wed all be dead by now
#6 2005-05-18 19:39:17
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Re: Last one hopefully
In other words ... Hitler/Maths would have had the power to defeat the Allies, and then would have gone on to kill everyone he now successfully ruled?
This is a new theory I hadn't encountered before.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#8 2005-05-19 07:30:30
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Re: Last one hopefully
So does anyone have any idea on how to do this problem?
#11 2005-05-19 21:32:52
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Re: Last one hopefully
He did ... I don't know what he replaced it with either.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#12 2005-05-20 01:34:36
- Roraborealis
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Re: Last one hopefully
A ferret reaper.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
Fatboy Slim is the band of the 90s, thats if you want to call it a band because its really a one man name #15 2005-05-21 08:51:53
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Re: Last one hopefully
well anyways back to my original question, I got some help from a friend and they said to start by making the (3/2)(x-1) into one term, and then multiply the sqrt by something that will allow you to combine it with the expanded fraction, then go from there. And they said that the final solution will end up with a square root inside of a square root, and that there wasnt a way that they knew to get around it.
#16 2005-05-21 20:03:06
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Re: Last one hopefully
Then, we'll first need to multiply out the brackets.
Is that 3/2 a three divided by 2?
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
#17 2005-05-22 05:28:20
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Re: Last one hopefully
Yes it is a 3 divided by a 2. Now after following some of their advice I combined both terms to make one term which is: (3x - 3) + (sqrt(2x^2 - 7x - 4)) All divided by 2. But I can't get any further at the moment
#18 2005-05-27 03:37:28
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Re: Last one hopefully
hey who ever u r i got the answer for u.telling u something tht go order wise and get ur answer.
the no not possibly will be a perfect sqr. so u can actaually do it calmly!ur answer + question is toooo difficult.
the answer by me is not valid.
NO VALUE.
COZ,
3/2 x -1 + sqrt of (2^2 -7 x -4)
=-3/2 + sqrt of 12
sqrt of 12 does not come in whole nos. u cant add approximation
( if u do it this way)
#19 2005-05-27 07:45:22
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Re: Last one hopefully
cool_me, you are genius ...
... I think 
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman