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 June 18th, 2012, 06:30 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Sum & Product of the Roots of a Quadratic Equation Having a bit of trouble with this one. can anyone help? Many thanks. Q. Find the value of p, if 1 root of the equation $3x^2+5x-p=0$ is 2 greater than the other root. Attempt: $a= 3, b = 5, c = -p$ Sum of Roots: $\alpha+2\alpha$ => $3\alpha$ => $\frac{-b}{a}$ => $\frac{-5}{3}$ => $3\alpha=\frac{-5}{3}$...(i) Product of Roots: $\alpha(2\alpha)$ => $2\alpha^2$ => $\frac{c}{a}$ => $\frac{-p}{3}$ => $2\alpha^2=\frac{-p}{3}$...(ii) From (i): $3\alpha=\frac{-5}{3}$ => $\alpha=\frac{-5}{9}$ From (ii): $2\alpha^2=\frac{-p}{3}$ => $2(\frac{-5}{9})^2=\frac{-p}{3}$ => $\frac{25}{81}=\frac{-p}{6}$ => $\frac{150}{81}=-p$ => $p=\frac{-50}{27}$ Ans.: (From text book): $p=\frac{11}{12}$
 June 18th, 2012, 07:37 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,659 Thanks: 964 Math Focus: Elementary mathematics and beyond Re: Sum & Product of the Roots of a Quadratic Equation $3x^2\,+\,5x\,-\,p\,=\,0\,\Rightarrow\,x^2\,+\,\frac53 x\,-\,\frac{p}{3}\,=\,0$ $(x\,-\,a)(x\,-\,(a\,+\,2))\,=\,x^2\,-\,(2a\,+\,2)x\,-\,a(a\,+\,2) \\ \Rightarrow\,-2a\,-\,2\,=\,\frac53 \\ \Rightarrow\,a\,=\,-\frac{11}{6},\,a\,+\,2\,=\,\frac16$ $-\frac{11}{6}\,\cdot\,\frac{1}{6}\,=\,-\frac{p}{3} \\ \frac{11}{36}\,=\,\frac{p}{3}\,\Rightarrow\,p\,=\, \frac{11}{12}$
June 18th, 2012, 07:41 AM   #3
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Re: Sum & Product of the Roots of a Quadratic Equation

Hello, bilano99!

Quote:
 $\text{Find the value of }p,\,\text{ if one root of the equation }\:3x^2\,+\,5x\,-\,p\:=\:0 \;\;\;\text{ is }\underbrace{\text{2 greater than the other root.}}_{\text{Here!}}$

$\text{The two roots are: }\:\alpha\,\text{ and }\,\alpha+2$

 June 18th, 2012, 08:02 AM #4 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Sum & Product of the Roots of a Quadratic Equation Ah! I see what I did. Thanks, guys.
 June 18th, 2012, 10:26 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Sum & Product of the Roots of a Quadratic Equation Another approach: The difference between the two roots is 2: $\frac{\sqrt{b^2-4ac}}{a}=2$ $\frac{\sqrt{5^2+12p}}{3}=2$ $25+12p=36$ $p=\frac{11}{12}$

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