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ok. How the hell do you apply the descriminant (B^2 - 4ac) to an equation like this:
x^2 + (k-2x)^2 - 3x - 4 = 0
with intentions of finding the value of k.
The question says that y+2x=k is tangent to circle x^2 + y^2 -3x - 4 = 0. And to find the value of k (7 marks) .
In the anwswer booklet it says "x^2 + (k-2x)^2 - 3x - 4 = 0 apply to descriminant"
HELP exam is tomorrow:|
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HElp day before exam!
Started by gsmushet1, May 18 2006 10:31 PM
5 replies to this topic
#2
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Posted 18 May 2006 - 10:36 PM
i would break the bracket and get k^2 - 4kx + 4x^2 this means we now have
5x2+(-4k-7)x+(k2-4)=0
a=5
b=-7-4k
c=k2-4
5x2+(-4k-7)x+(k2-4)=0
a=5
b=-7-4k
c=k2-4
If i am not here i am somewhere else
#3
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Posted 18 May 2006 - 10:37 PM
Expand the (k-2x)² bracket and simplify the expression, separate the x² terms, x and numberical terms. k is a number so if you get that on it own or a multiple of that without x in the value, just separate it, and group like terms together. I'd go in depth, but someone's bound to post soon enough anyway! 
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#4
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Posted 18 May 2006 - 10:53 PM
QUOTE(Dave @ May 18 2006, 11:36 PM) 
i would break the bracket and get k^2 - 4x + 4x^2 this means we now have
5x2-7x+(k2-4)=0
a=5
b=-7
c=k2-4
but if you break the bracket do you not get k^2 - 4kx + 4x^2
#5
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Posted 18 May 2006 - 11:04 PM
yeah u do doesnt change anything though
a=5
b=-7-4k
c=k2-4
a=5
b=-7-4k
c=k2-4
If i am not here i am somewhere else
#6
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Posted 19 May 2006 - 01:51 PM
Ach nevermind, it's over now and i'm SURE of an A=)
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