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805+ Level|   Absolute Values|   Algebra|      
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 03:54
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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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B
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 05:32
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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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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 04:50
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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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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 05:26
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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}
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Bunuel
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


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 [ 1 is subtracted because negative of 0 is also 0.]
          • That is p can be {0,10,18,24,28,30,-10,-18,-24,-28,-30}
Thus, the correct answer is Option B.
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 06:21
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See the attachment,
My Answer is B.
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1.PNG [ 49.33 KiB | Viewed 6142 times ]

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Bunuel
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


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
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 09:52
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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.
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 29 May 2020, 12:05
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Bunuel
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

\(x^2 - 11x + |p| = 0\) 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

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B
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 30 May 2020, 02:47
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Bunuel
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

\(x^2 - 11x + |p| = 0\) 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

Show Spoiler
B

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.
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If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has 19 Jan 2021, 03:59
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hi, why cant be roots as (16,-5) in that case too sum of the roots will be 11 ...
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AAspirations
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.

\((x+1)(x-12) = x^2 - 11x - 12\)
It is not possible that \(x^2 - 11x + |p| = x^2 - 11x - 12\), since |p| cannot be equal to -12.
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vipulgoel
hi, why cant be roots as (16,-5) in that case too sum of the roots will be 11 ...

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