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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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nick1816
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x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   19 Jun 2019, 16:43
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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
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C
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] 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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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] 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] New post   09 Dec 2019, 05:58
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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


(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)
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x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   25 Sep 2020, 03:42
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



Bunuel VeritasKarishma

Could you please elaborate on this question.
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   25 Sep 2020, 04:35
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Standard quadratic equation is: \({ax}^2\) + bx + c = 0

Given equation: \(x^2\) + bx - c = 0 [a = 1 ; b = b ; c = -c]

Roots: 'u' and 'v'

Sum of roots: \(\frac{- b }{ a }\)

=> u + v = \(\frac{-b }{ 1}\) = - b

=> Some of the two numbers are negative when [both are negative or anyone is more negative than others.]

Product of roots: \(\frac{c }{ a}\)

=> u * v = \(\frac{-c }{ a}\) = -c

=> Product of two numbers is negative when [both have opposite sign]

Since, lul > lvl, it is only possible when u is more negative than v.

Therefore, u < 0 and v > 0.

Answer C
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   25 Sep 2020, 05:26
MathRevolution wrote:
Standard quadratic equation is: \({ax}^2\) + bx + c = 0

Given equation: \(x^2\) + bx - c = 0 [a = 1 ; b = b ; c = -c]

Roots: 'u' and 'v'

Sum of roots: \(\frac{- b }{ a }\)

=> u + v = \(\frac{-b }{ 1}\) = - b

=> Some of the two numbers are negative when [both are negative or anyone is more negative than others.]

Product of roots: \(\frac{c }{ a}\)

=> u * v = \(\frac{-c }{ a}\) = -c

=> Product of two numbers is negative when [both have opposite sign]

Since, lul > lvl, it is only possible when u is more negative than v.

Therefore, u < 0 and v > 0.

Answer C


Why can't v be a small positive and u be a big negative number? That way, u+v= -ve; u*v=-ve; |u|>|v|. All three conditions met, and u and v are opposite signed too.
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   27 Sep 2020, 23:52
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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



\(x^2 + bx – c = 0\)

Sum of the roots = -b (a negative number since b is positive)

Product of the roots = -c (a negative number since c is positive. Not 0 so either root cannot be 0)
This means one root is positive and the other negative.

Since sum is negative too, it means the negative number has a greater absolute value. Since absolute value of u is greater than absolute value of v, u is negative and v is positive.

Answer (C)
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   03 Dec 2020, 22:14
MathRevolution wrote:
Standard quadratic equation is: \({ax}^2\) + bx + c = 0

Given equation: \(x^2\) + bx - c = 0 [a = 1 ; b = b ; c = -c]

Roots: 'u' and 'v'

Sum of roots: \(\frac{- b }{ a }\)

=> u + v = \(\frac{-b }{ 1}\) = - b

=> Some of the two numbers are negative when [both are negative or anyone is more negative than others.]

Product of roots: \(\frac{c }{ a}\)

=> u * v = \(\frac{-c }{ a}\) = -c

=> Product of two numbers is negative when [both have opposite sign]

Since, lul > lvl, it is only possible when u is more negative than v.

Therefore, u < 0 and v > 0.

Answer C


I get the logic behind using Viete's theorem, but what I don't get is the signs of b and c.

As far as I can tell you only care about "b" and "c" and not the sign that is in front in the quadratic equation.
Even if that is not the case, we see here that b is positive and c is negative so why make b negative like above?

Can someone clarify the signs that we should focus on from the quadratic?
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Re: x^2 + bx – c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   14 Dec 2020, 12:52
can someone explain why i'm wrong?

here's how i approached:

since U & V are roots of the equation:

substituting U in the equation

U^2 + U . B - C = 0

taking C on the right hand side

U^2 + U . B = (positive C)

Since U^2 can't be negative and given that B > 0

Only case for positive U > 0
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Re: x^2 + bx c = 0; b and c are positive integers. u and v are roots of [#permalink] New post   08 Sep 2023, 11:13
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

Hence, OA C
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Re: x^2 + bx c = 0; b and c are positive integers. u and v are roots of [#permalink]
08 Sep 2023, 11:13
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