Solving a Quadratic Equation: Find Real Values of 'p

  • #1
29
1
Hi,
I may not be thinking straight or something, and I am having trouble with this question, please help!:

Given that 'p' is real, find the set of values of 'p' for which the roots of:
(2p+1)x^2 - 10x + p-2 = 0
a) Are real
b) Have a sum>5

Im thinking along the lines of: alpha+beta = -b/a and alpha x beta = c/a however I end up with 10/(2p+1) and (p-2)/(2p+1) respectively, which I can't work out simultaneously (2 equations for 3 variables).. Maybe I am making something really stupid but please help.

Thanks a lot for all your help.
 
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  • #2
HINT: (a) findthe roots using the quadratic formula.

(b) use the discriminant for the first condition.

(c) add the roots for the second condition.
 
  • #3
Ahh thanks a ton! Don't know why I ignored those possibilities...
Whilst we are on the topic, hehe, how would you draw 4<(x-4)^2 + (y-3)^2 <25 on a number plane? I can draw it on a cartesian graph using (x-4)^2 + (y-3)^2 = 25 and (x-4)^2 + (y-3)^2 = 4 but am confused as to how to transform this onto a simple number plane.. any hints there?
 
  • #4
That looks like a circle centered on (4, 3) and its radius is between 2 and 5.
 

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