ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lacoste on February 15, 2009, 04:42:40 pm
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ahow that the equation (k+1)x^2-2x-k=0 has a solution for all values of k.
thanks~!!!
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Is the equation supposed to read
^2-2x-k=0 )
or
x^2 -2x-k =0 )
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By 'a' solution I'm assuming 1 or more solutions.
The discriminant
can be used to find the number of solutions. If
, then the equation will have solutions.
^2 - 4(k+1)(-k) )
 )

 )
)
^2 + \frac{3}{4}\right) )
^2 + 3)
Since
, we can see that
. So the equation will always have 2 solutios for any value of k.
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thanks divide 0!!
cobby: the second one the x^2
Is the equation supposed to read
^2-2x-k=0 )
or
x^2 -2x-k =0 )
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hey divide0, did you accidently place 3 instead of +3/4 at the end of the working? or did something cancel out which i dont know where?
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Before he completed the square, he took 4 out of the equation as a factor.
When hes completing the square, the 4 still remains outside as a factor.
so the
cancels down to 3.
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thanks dominicz.
get it now. 12/4=3 cheers