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

Question

If  x^2 - 11x + |p| = 0 , where x is a variable and p is a constant, has two distinct integer roots, then how many integer values can p take?

A. 12
B. 11
C. 10
D. 9
E. 8

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prateek25 · Accepted Answer

Let the roots be u, t

u + t = sum of the roots = -b/a = 11
ut = product of the roots = c/a = |p|

Equation 1 above gives 12 pairs of u,t 
0,11; 1,10 ; 2,9; 3,8; 4,7; 5,6; 6,5; 7,4, 8,3; 9,2 , 10,1; 11,0

Only 6 of these above have distinct products.
So there are 6 possible values of |p|.

But since p can be positive or negative for |p| to be positive, there are 5 additional values of p possible (p=0 will not have a negative counterpart).

Total possible values of p= 11

Answer should be B

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yashikaaggarwal · Answer

IMO B

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)

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NitishJain · Answer

B. 11

For the equation to have integer roots, the  discriminant  (D=b^2 - ac) must be greater or equal to zero AND must be a perfect square

For integer roots: b^2–4ac≥0
b= 11
a=1
c= |p|

==> 11^2 -4|p| ≥ 0
==> 121 ≥ 4|p|
or |p| ≤ 121/4
or |p| ≤ 30.25
or -30.25 ≤ p ≤ 30.25

for  discriminant   to be perfect values p = {-30, -28, -24, -18, -10, 0 ,10, 18, 24, 28, 30}

GMATWhizTeam · Answer

Solution 
 • Let us assume that a and b are two distinct integral roots of the equation,  x^2 -11x + |p| = 0 
 o Therefore,  a + b = -\frac{(11)}{1} = 11 
o And,  a*b = \frac{|p|}{1} =|p | 
• From above we can see that summation of  a + b > 0  and  a*b ≥0 
 o Therefore, both a and b must be non-negative. 
o Since, a and b are distinct integers, so possible values of (a,b) = (0,11), (1,10),(2,9), (3,8), (4, 7), and (5, 6)
o Correspondingly, |p| can be (0*11), (1*10), (2*9),(3*8),(4*7), and (5*6)
  Total 6 integral values are possible for |p|
 Since, |p| = 10 implies that p can be either 10 or -10.
 • Therefore, number of possible integer values of p will be 6*2 – 1 = 11  
• That is p can be {0,10,18,24,28,30,-10,-18,-24,-28,-30}    
Thus, the correct answer is  Option B.

sambitspm · Answer

See the attachment,
My Answer is B.

GMATinsight · Answer

PROPERTY  : 
For a quadratic equation given by  y = ax^2+bx+c 
Sum of the roots =  \frac{-b}{a} 
Product of the roots =  \frac{c}{a}

Here sum of the roots = -(-11)/1 = 11

i.e. Sum of two integers is 11 while product of them should be a Non-Negative Integer lpl because Product of the roots = lpl

11 = 
  0+11 i.e. lpl = 0 
1+10 i.e. lpl = 10 i.e. p = +10 or -10
2+9 i.e. lpl = 18 i.e. p = +18 or -18
3+8 i.e. lpl = 24 i.e. p = +24 or -24
4+7 i.e. lpl = 21 i.e. p = +21 or -21
5+6 i.e. lpl = 30 i.e. p = +30 or -30
  
 i.e. 11 possible solutions of p

Answer: Option B

naveenban · Answer

If the Roots are integers, ( −11±√D ) / 2 ( Roots of the equation ) have to be integers.
For the above value to be an integer, D has to be a perfect square and an odd number, so that the resultant is an Integer too ( Divided by 2 ).

D = b^2 - 4ac = 121 – 4|p| 
4|p| cannot be negative => D can take values 121, 81, 49, 25, 9, 1
|p| can be 0, 10, 18, 24, 28, 30
p can take 0, ±10, ± 18, ± 24, ± 28, ± 30 = 11 Values

Hence, answer option B.

GMATGuruNY · Answer

x^2 - 11x + |p| = 0  can be factored as follows:
(x - a)(x - b), where a and b are nonnegative integers such that a+b = 11 and ab = |p|.

Options:
(x - 0)(x - 11) --> |p| = 0*11 = 0 --> p=0
(x - 1)(x - 10) --> |p| = 1*10 = 10 --> p=±10
(x - 2)(x - 9) --> |p| = 2*9 = 18 --> p=±18
(x - 3)(x - 8) --> |p| = 3*8 = 24 --> p=±24
(x - 4)(x - 7) --> |p| = 4*7 = 28 --> p=±28
(x - 5)(x - 6) --> |p| = 5*6 = 30 --> p=±30
Number of options for p = 11

B

AAspirations · Answer

Why should a and b be nonnegative integers. It doesn't say they have to be. You could have (x+1)(x-12) for instance. Please clarify.

vipulgoel · Answer

hi, why cant be roots as (16,-5) in that case too sum of the roots will be 11 ...

GMATGuruNY · Answer

(x+1)(x-12) = x^2 - 11x - 12 
It is not possible that  x^2 - 11x + |p| = x^2 - 11x - 12 , since |p| cannot be equal to -12.

GMATGuruNY · Answer

The product of the two roots = |p| = NONNEGATIVE.
Thus, the two roots cannot have different signs.

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