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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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VP
Joined: 19 Oct 2018
Posts: 1294
Location: India
x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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19 Jun 2019, 16:43
1
10
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Difficulty:

75% (hard)

Question Stats:

49% (02:08) correct 51% (02:08) wrong based on 77 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
VP
Joined: 20 Jul 2017
Posts: 1247
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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19 Jun 2019, 17:38
5
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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##### General Discussion
Intern
Joined: 16 Dec 2018
Posts: 5
Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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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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VP
Joined: 24 Nov 2016
Posts: 1094
Location: United States
Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of  [#permalink]

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09 Dec 2019, 05:58
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

(u,v) = roots; |u|>|v|;
b and c = positive integers;

vietas formula sum roots: u+v = -b/a = -b/1 = -b
u+v=-b: one must be more negative than the other, or both negative.

vietas formula product roots: uv = c/a = (-c)/1 = -c
uv=-c: (u,v) are not 0 and they have different signs.

(u,v) have different signs and one is larger than the other so that their sum is a negative.

Ans (C)
Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of   [#permalink] 09 Dec 2019, 05:58
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