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July 23, 2014

July 23, 2014

Posted by **Rikkie** on Monday, June 15, 2009 at 10:17pm.

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

simplify....

x^2 + (2p+1)x + p^2 = 0

b^2-4ac > 0

(-2p-1)^2 - 4(p^2) > 0

simplify....

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 10:45pmactually 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 10:56pmthanks!

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