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x^2 + bx – c = 0; b and c are positive integers. u and v are roots of

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Director
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Joined: 19 Oct 2018
Posts: 671
Location: India
x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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New post 19 Jun 2019, 16:43
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Difficulty:

  65% (hard)

Question Stats:

47% (02:00) correct 53% (01:41) wrong based on 38 sessions

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\(x^2 + bx – c = 0\); b and c are positive integers. u and v are roots of the quadratic equation . If |u| > |v|, then which of the following is true?

A. u>0 and v>0
B. u>0 and v<0
C. u<0 and v>0
D. u<0 and v<0
E. u=0 and v<0
Director
Director
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Joined: 20 Jul 2017
Posts: 521
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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New post 19 Jun 2019, 17:38
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nick1816 wrote:
\(x^2 + bx – c = 0\); b and c are positive integers. u and v are roots of the quadratic equation . If |u| > |v|, then which of the following is true?

A. u>0 and v>0
B. u>0 and v<0
C. u<0 and v>0
D. u<0 and v<0
E. u=0 and v<0


Formula: For the quadratic equation, ax^2 + bx + c = 0.
Sum of the roots = -b/a
Product of the roots = c/a

For the given equation, x^2 + bx- c = 0

Sum of the roots, u + v = -b/1 = -b
—> Both are negative or at least 1 is negative

Product of the roots, u*v = -c/1 = -c
—> u and v are of opposite signs and
since lul > lvl,
—> u is more negative and v is positive {Eg: (u, v) = (-3, 1), (-10, 8) etc}
—> u < 0 and v > 0

IMO Option C

Pls Hit kudos if you like the solution

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Intern
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Joined: 16 Dec 2018
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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New post 27 Jun 2019, 10:15
I am not sure this might be a bit confusing for you. But let try:

'b' should be factorise in such a way that
-->product of it's factors should be negative (as c is negative)
but
-->sum should be positive (as b is positive)

When sum should be positive than 'b'should be factorise in a way which have bigger one positive number and other smaller negative number

So when taken as roots bigger magnitude number become negative root and smaller magnitude number become positive root

As it is given magnitude of 'u' is grater than 'v'; we can say 'u' is smaller than 0 and 'v' is greater than 0

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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of   [#permalink] 27 Jun 2019, 10:15
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x^2 + bx – c = 0; b and c are positive integers. u and v are roots of

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