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DS: X^2+ BX+ C=0,---> R,S are solutions of the

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New post 07 Sep 2004, 07:24
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DS: X^2+ BX+ C=0,---> R,S are solutions of the equation.Is R* S>0?
1. B<0
2. C<0
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New post 07 Sep 2004, 07:49
B.

Since R and S are solutions of the equation, the equation can be written as x^2 - (R+S)x + RS = 0

In this case, B = -(R+S) and C = RS.
From II, C < 0 => RS < 0, Ans to the question: No.
Joined: 31 Dec 1969
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New post 07 Sep 2004, 08:59
For the roots to be real and distinct in a quadratic equation, the determinant b^2-4ac > o

here b = B
c=C
a = 1

so, w have B^2 - 4C > 0

Now, 1) B < 0. Doesn't matter as B^2 is always +ve. So we dont know if B^2- 4C will be +ve or not. So insufficient

2) C<0. Sufficient. The B^2 - 4C will be > than 0.

Hence B.
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New post 07 Sep 2004, 09:00
guest was me..sorry forgot to login earlier.
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New post 07 Sep 2004, 12:13
Great work guys!

I took C as answer, while now it is obvious for me as well that B is sufficient. We need to understand the value of C to answer the question.

Regards,

Alex
  [#permalink] 07 Sep 2004, 12:13
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DS: X^2+ BX+ C=0,---> R,S are solutions of the

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