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

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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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Bunuel wrote:
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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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Bunuel wrote:
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

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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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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

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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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Bunuel wrote:
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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
GMATGuruNY wrote:
Bunuel wrote:
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

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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
hi, why cant be roots as (16,-5) in that case too sum of the roots will be 11 ...
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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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AAspirations wrote:
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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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vipulgoel wrote:
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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Re: If x^2 - 11x + |p| = 0, where x is a variable and p is a constant, has [#permalink]
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