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x^2 − 11x + |p| = 0, where x is a variable and p is a constant, has 2

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KarunMendiratta
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x^2 − 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   24 Sep 2021, 10:51
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x2−11x+|p|=0, where x is a variable and p is a constant, has 2 distinct integer roots.

Quantity A
Quantity B
Number of integer values that p can take
10


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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motion2020
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x^2 − 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   24 Sep 2021, 13:55
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very typical of the discriminant case problem

√112−4∗|p|>0 and the discriminant actually must be odd integer for x to have integer roots

there are not many perfect squares after the multiples of 4 are subtracted; one is 92 and |p|=8

another is 72 and |p|=18

yet, another is 52 and |p|=24
...

summing up, all perfect squares of odd numbers below 11 and including 11, i.e. 11, 9, 7, 5, 3, 1 are suitable here

p={-30, -28, -24, -18, -8, 0, 8, 18, 24, 28, 30} set contains 11 integers

answer is A, as Quantity A=11 > Quantity B=10



KarunMendiratta wrote:
x2−11x+|p|=0, where x is a variable and p is a constant, has 2 distinct integer roots.

Quantity A
Quantity B
Number of integer values that p can take
10


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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AnkitaShinde
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   22 Aug 2022, 12:44
Is there some other solution?
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   22 Aug 2022, 12:59
AnkitaShinde wrote:
Is there some other solution?


There are several more solutions to this question on GMAT Club (GRE Prep Club's sister site): https://gmatclub.com/forum/if-x-2-11x-p ... 25456.html
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   20 Dec 2023, 03:09
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This is my own solution


For the equation to have 2 distinct roots there are two cases

Case 1


If p is positive


b^2-4ac>0

121-4p>0

-4p>-121

p<121/4

Case 2

Where p is negative


121+4p>0

4p>-121

p>-121/4


Combining both inequalities we have

-121/4<p<121/4

-30.25<p<30.25


Integer values will then be

-30<=p<=30



Of these p can take integers -30, -28, -24, -18, -10, 0, 10, 18, 24, 28, 30.

Quantity A is greater

Motion2020 made a mistake with -8 and 8, 10 and -10 will go instead. Hope it helps.

Adewale Fasipe, GRE/GMAT quant instructor from Nigeria. You may contact me for classes here.
https://www.teacheron.com/tutor-profile/8tG7?r=8tG7
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink] New post   05 Feb 2025, 03:16
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink]
05 Feb 2025, 03:16
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