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June 3rd, 2011, 02:38 AM   #1
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Quadratics

Consider the quadratic equation 2px+(p-1)x+2p=0

a. Find the discriminant
b. Find the values of p for which there are 2 solutions
c. Find the value of p for which there are no solutions
d. Find the value of p for which there is 1 solution

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June 3rd, 2011, 03:06 AM   #2
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Re: Quadratics

a. (p-1)^2-16p^2 = (p-1-4p)(p-1+4p)=-(3p+1)(5p-1)

p = -1/3 , p= 1/5
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June 3rd, 2011, 04:27 AM   #3
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Re: Quadratics

Think of the quadratic formula:

Given then



The term under the radical is termed the discriminant.

When the discriminant is greater than zero, there are two real solutions.

When the discriminant is equal to zero, there is one real solution (a repeated root of multiplicity two).

When the discriminant is less than zero, there are two complex conjugate roots (no real solution).

As subhee has shown, the discriminant may be expressed as -(3p + 1)(5p - 1) which has zeroes of -1/3 and 1/5. So plot those points on a number line, creating the open intervals (where we may choose a test point from each interval):

(-?,-1/3) test point -1: -(-)(-) = -, no real solutions on this interval

(-1/3,1/5) test point 0: -(+)(-) = +, 2 real solutions on this interval

(1/5,?) test point 1: -(+)(+) = -, no real solutions on this interval
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June 4th, 2011, 01:58 AM   #4
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If y = ax + bx + c where a ? 0, b and c are real numbers, then the discriminant ? = b - 4ac.
If ? > 0, there are two solutions.
If ? = 0, there is one solution.
If ? < 0, there are zero solutions.
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