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HElp day before exam! - HSN forum

# HElp day before exam!

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### #1gsmushet1

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Posted 18 May 2006 - 10:31 PM

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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:|

### #2Dave

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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

If i am not here i am somewhere else

### #3The Wedge Effect

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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!

### #4gsmushet1

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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

### #5Dave

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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

If i am not here i am somewhere else

### #6gsmushet1

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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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