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EgmatQuantExpert

Solution


Given:
    • p > 0
    • \(x^2 – 11x + p = 0\) has integer roots

To find:
    • Number of possible integral values of p

Approach and Working:
Discriminant must be a perfect square. Since, a perfect square cannot be negative
    • \(b^2 – 4ac ≥ 0\)
    • \(11^2 – 4p ≥ 0\)
    • Thus, p ≤ 30.25

Therefore, 0 < p ≤ 30.25

Perfect squares in this range are {1, 4 , 9, 16, 25}

Hence, the correct answer is Option C

Answer: C


This answer is wrong though. None of those solutions {1,4, 9, 16, 25} are correct for the problem.
x^2 -11x+8 has positive p roots at (-10,-1) (-9, -2) (-8, -3) (-7, -4) (-6, -5) or P={10, 18, 24, 28, 30}
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EgmatQuantExpert

Solution


Given:
    • p > 0
    • \(x^2 – 11x + p = 0\) has integer roots

To find:
    • Number of possible integral values of p

Approach and Working:
Discriminant must be a perfect square. Since, a perfect square cannot be negative
    • \(b^2 – 4ac ≥ 0\)
    • \(11^2 – 4p ≥ 0\)
    • Thus, p ≤ 30.25

Therefore, 0 < p ≤ 30.25

Perfect squares in this range are {1, 4 , 9, 16, 25}

Hence, the correct answer is Option C

Answer: C


This answer is wrong though. None of those solutions {1,4, 9, 16, 25} are correct for the problem.
x^2 -11x+p has positive p roots at (-10,-1) (-9, -2) (-8, -3) (-7, -4) (-6, -5) or P={10, 18, 24, 28, 30}
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cfc198
If p > 0, and x^2 - 11x + p = 0 has integer roots, how many integer values can 'p' take?

A,6
B.11
C.5
D.10
E.0

If you liked the question, please do hit the kudos button :)

Asked: If p > 0, and x^2 - 11x + p = 0 has integer roots, how many integer values can 'p' take?

If a & b are roots of the equation x^2 - 11x + p = 0 & p> 0 so a & b are both positive;

a + b = 11
(a , b) = {(1,10),(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2),(10,1)}
ab = p
p = {10,18,24,28,30}

IMO C
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Can anybody explain why re-writing the equation as p=11x-x^2 and then finding the solution lead to an incorrect answer?

just curious to see why this method is ineffective..
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mosal101
Can anybody explain why re-writing the equation as p=11x-x^2 and then finding the solution lead to an incorrect answer?

just curious to see why this method is ineffective..

This approach will yield the correct answer.

p = x(11-x)

Since it is given that the equation has integer roots, p will be positive only if 1≤x≤10, yielding the following options:
x=1 and 11-x=10 --> p = 1*10 = 10
x=2 and 11-x=9 --> p = 2*9 = 18
x=3 and 11-x=8 --> p = 3*8 = 24
x=4 and 11-x=7 --> p = 4*7 = 28
x=5 and 11-x=6 --> p = 5*6 = 30
x=6 and 11-x=5 --> p = 6*5 = 30
x=7 and 11-x=4 --> p = 7*4 = 28
x=8 and 11-x=3 --> p = 8*3 = 24
x=9 and 11-x=2 --> p = 9*2 =18
x=10 and 11-x=1 --> p = 10*1 = 10

Resulting options for p:
10, 18, 24, 28, 30
Total number of options = 5

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