Posted by Rikkie on Monday, June 15, 2009 at 10:17pm.
This is the question:
The straight line y=2p+1 intersects the curve y= x+ (p^2/x) at two distinct points. Find the range of values of p.
this is what i did:
2p +1 = x + p^2/x
x^2 + (2p+1)x + p^2 = 0
b^2-4ac > 0
(-2p-1)^2 - 4(p^2) > 0
4p + 1 > 0
P > -1/4
However, the correct answer should be:
p>-1/4, p is not equals to 0.
How do i get the second part "p not equals to 0"??
Please check my maths answer? - Reiny, Monday, June 15, 2009 at 10:45pm
actually your equation should have been
x^2 - (2p+1)x + p^2 = 0
but since you are squaring b in
b^2 - 4ac > 0 it did not show up in your solution.
the restriction 'p not equal to zero' shows up when we analyse our equations
when p=0 the first equation is the horizontal line y = 1 and the second equation is no longer a curve, but rather simply y = x
these clearly intersect at ONE point (1,1), not at two.
since p = 0 falls in the domain of p > -1/4, we have to restrict it.
thanks! - Rikkie, Monday, June 15, 2009 at 10:56pm
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