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Find the value of the values of k such that the equaltion x(squared) + 4kx + 3k = 0 has two distinct roots!
iv got it down to 4k(4k-3)>0 so therefore k> 3/4 but whats the other one?
thanks
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Posted: 2006-10-01 20:19:06
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zero?
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Posted: 2006-10-01 21:35:07
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thats what i thought aswell cheers
can you solve the equation 3x(squared) + 5x + 8?
cheers
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Posted: 2006-10-01 21:38:25
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im not sure thats right anymore, since zero would suggest it only had one real root, how did you get to the line you wrote?
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Posted: 2006-10-01 21:40:19
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b(squared) - 4ac
(4k)squared - 4(1x3k) >0
16k(squared) - 12k >0
4k(4k-3) >0
thats a good point
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Posted: 2006-10-01 21:44:14
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You're doing it right till 4k(4k - 3) > 0
Now for this to be possible k can have values from (-infinity to 0) and (3/4 to infinity)
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Posted: 2006-10-02 00:31:28
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hey, any of you guys use Mathematica soft? I pretty fancy it recently:)
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Posted: 2006-10-04 12:31:51
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I used Mathematica 4.1 and Mathematica 5 when I was in high school. I don't remember exacly but it was some Wolfram software
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Posted: 2006-10-08 22:26:00
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-b(+ or -)root of (b(squared) -4AC)
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2A
a b and c being (in example of your equation) A(squared) + Bx + C, i dunno how to put it really...
(+ or - as there is usually two solutions, do + AND - separetly)
put your values in and it should give you the answer
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[ This Message was edited by: Glenny on 2006-10-08 22:23 ]
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Posted: 2006-10-08 23:22:41
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lol it was easy. K=1 using an algebra factorization method.
Done in paint. Sorry for the quality. In the end, the ecuation results zero, so either factor must be zero, giving us two roots. Cheers
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Posted: 2006-10-09 07:15:50
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