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 June 20th, 2012, 06:47 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 me out? Many thanks. Q. For what value of m are the roots of the equation $3x^2+(m-1)x-2=0$ equal but opposite in sign. Attempt:$a=3,b=m-1,c=-2$ Sum of roots: $\alpha-\alpha=0$ => $\frac{-b}{a}$ => $\frac{-(m-1)}{3}$ => $\frac{-m+1}{3}=0$ Product of roots: $\alpha(-\alpha)$ => $-\alpha^2$ => $\frac{c}{a}$ => $\frac{-2}{3}$ => $-\alpha^2=\frac{-2}{3}$ => $\alpha^2=\frac{2}{3}$ => $\alpha=\sqrt{\frac{2}{3}}$ $\sqrt{\frac{2}{3}}=\frac{-m+1}{3}$ => $\frac{2}{3}=\frac{m^2-2m+1}{9}$ => $18=3m^2-6m-15$ => $3m^2-6m-15=0$ => $m^2-2m-5=0$ => $(m-1)(m-5)$... Ans.: (From text book): m = 1
 June 20th, 2012, 06:59 AM #2 Member   Joined: Jun 2012 From: UK Posts: 39 Thanks: 0 Re: Sum & Product of the Roots of a Quadratic Equation You were virtually there, and then wandered off into deep doodoo. Just considering the sum of the roots gets you to (-m+1)/3=0 and hence m=.... The only other thing you need to consider is that the product of the roots is non-zero (otherwise both roots are zero), and that's true, so you're done.
June 20th, 2012, 07:12 AM   #3
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Re: Sum & Product of the Roots of a Quadratic Equation

Hello, bilano99!

You had the answer . . . You just zoomed past it.

Quote:
 $\text{For what value of }m\text{ are the roots of: }\:3x^2\,+\,(m-1)x\,-\,2\:=\:0 \;\;\;\text{ equal but opposite in sign.}$

$\text{W\!e have: }\:x^2\,+\,\frac{m-1}{3}x\,-\,\frac{2}{3} \:=\:0$

$\text{Let the two roots be }\alpha\text{ and }-\alpha.$

$\text{Then: }\:\begin{Bmatrix}\text{sum of the roots: }=&\alpha\,+\,(-\alpha) \:=\:0 \\ \\ \\ \text{sum of the roots: } &-\frac{m\,-\,1}{3} \end{Bmatrix}=$

$\text{Therefore: }\:-\frac{m\,-\,1}{3} \:=\:0 \;\;\;\Rightarrow\;\;\;m \:=\:1$

 June 20th, 2012, 07:34 AM #4 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Sum & Product of the Roots of a Quadratic Equation D'oh!!! Thanks, guys.
 June 20th, 2012, 08:32 PM #5 Global Moderator   Joined: Dec 2006 Posts: 17,729 Thanks: 1360 If the roots are p and -p, where p > 0, the equation is effectively 3(x + p)(x - p) = 0, i.e. 3x² - 3p² = 0, so m = 1.

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