June 3rd, 2011, 02:38 AM  #1 
Newbie Joined: Nov 2010 Posts: 27 Thanks: 0  Quadratics
Consider the quadratic equation 2px²+(p1)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 Thanks! 
June 3rd, 2011, 03:06 AM  #2 
Newbie Joined: May 2011 Posts: 9 Thanks: 0  Re: Quadratics
a. (p1)^216p^2 = (p14p)(p1+4p)=(3p+1)(5p1) p = 1/3 , p= 1/5 
June 3rd, 2011, 04:27 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  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 
June 4th, 2011, 01:58 AM  #4 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
If y = ax² + bx + c where a ? 0, b and c are real numbers, then the discriminant ? = b²  4ac. If ? > 0, there are two solutions. If ? = 0, there is one solution. If ? < 0, there are zero solutions. 

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