Algebra Pre-Algebra and Basic Algebra Math Forum

 June 3rd, 2011, 03:38 AM #1 Newbie   Joined: Nov 2010 Posts: 27 Thanks: 0 Quadratics Consider the quadratic equation 2px²+(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 Thanks!
 June 3rd, 2011, 04:06 AM #2 Newbie   Joined: May 2011 Posts: 9 Thanks: 0 Re: Quadratics a. (p-1)^2-16p^2 = (p-1-4p)(p-1+4p)=-(3p+1)(5p-1) p = -1/3 , p= 1/5
 June 3rd, 2011, 05:27 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Quadratics Think of the quadratic formula: Given $ax^2+bx+c=0$ then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ The term under the radical $b^2-4ac$ 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, 02:58 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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