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The set of values of $p$ for which the roots of the equation $3x^2+2x+p(p–1) = 0$ are of opposite sign is
 

  1. $(–∞, 0)$
  2. $(0, 1)$
  3. $(1, ∞)$
  4. $(0, ∞)$
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2 Answers

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Roots of equation are of opposite sign then for the equation, $ax^2+bx+c =0$,

product of roots, $\dfrac{c}{a}$, should be (negative number) less than $0$

$\dfrac{p(p-1)}{3}< 0$

$p(p-1)<0$

so $p$ must be less than $1$ and greater than $0$

Option B

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Let the roots are C and -D.

So we get C-D=2/3

and C*D=P(P-1)/3 should be <0 as the product of Cand D is negative as they are opposite signs.

P(P-1)/3<0

p(P-1)<0 WHICH INDICATE S THE VALUE SHOULD LIE  between 0 and 1 and the option B is correct.
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