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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink]
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]
1
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.
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Re: x^2 11x + |p| = 0, where x is a variable and p is a constant, has 2 [#permalink]
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]
Expert Reply
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]
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