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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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\(x^2 − 11x + |p| = 0\), 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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A
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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

\(\sqrt{11^2-4*|p|}>0\) 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 \(9^2\) and \(|p|=8\)

another is \(7^2\) and \(|p|=18\)

yet, another is \(5^2\) 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:
\(x^2 − 11x + |p| = 0\), 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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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   30 Jun 2025, 04:13
Can anyone explain me why p takes values as -30, -28, -24, -18, -10, 0, 10, 18, 24, 28, 30?
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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   30 Jun 2025, 04:17
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For two distinct roots D>0
D = b^2-4ac
D = 11^2-4*1*P
D = 121-4|P|
Let D = 0
121 = 4P => P = 30.25 or P<30.25

Since P should be an integer.
And 11 can't be factorized.
There middle term split will be done either in subtraction or addition
Such as 11= (11,0),(10,1)(9,2)(8,3)(7,4)(6,5)....
Putting value of P = 0,10,18,24,28,30(maximum value)
Putting values in equation x^2-11x+|p|, we get....
=>P=0 => X^2-11x=0 => X(x-11)=0 =>x=0 and X=11

=> P=10 => X^2-11x+10=0
=> X^2-10x-x+10
=> X(x-10)-1(x-10)
=> (x-10)(x-1) => X=10&1.....
Similarly,
(X-9)(X-2), (X-8)(X-3),
(X-7)(X-4), (X-6)(X-5),
All will be satisfying the equation.
So the maximum value P can take is 6, but since P is in modulus.
|P| will take opposite sign of +P as well. Such as {-30,-28,-24,-18,-10,0,10,18,24,28,30}
Total 11 (since Zero can't be positive or negative)

A is the answer
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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   16 Oct 2025, 07:19
Why are we only considering even values for p in the interval - -30<=p<=30 ?



[quote="adewale223"]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.
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink]
16 Oct 2025, 07:19
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