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# If p > 0, and x2 - 11x + p = 0 has integer roots, how many integer val

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Manager
Joined: 29 Nov 2018
Posts: 138
Concentration: Marketing, Strategy
If p > 0, and x2 - 11x + p = 0 has integer roots, how many integer val  [#permalink]

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01 Jan 2019, 11:35
2
1
2
00:00

Difficulty:

65% (hard)

Question Stats:

49% (02:02) correct 51% (02:03) wrong based on 47 sessions

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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

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Manager
Joined: 15 Jan 2018
Posts: 182
Location: India
Concentration: General Management, Finance
GMAT 1: 720 Q50 V37
WE: Information Technology (Computer Software)
Re: If p > 0, and x2 - 11x + p = 0 has integer roots, how many integer val  [#permalink]

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01 Jan 2019, 13:20
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Oh boy, such a difficult and time consuming question. Expect such PS question only if you're doing exceedingly well in the Quant section

Solution:
Condition 1: p > 0, and
Condition 2: x^2 - 11x + p = 0 has integer roots

We know that for a quadratic equation ax^2 + bx + c = 0, sum of the roots is (-b/a) and product of roots is (c/a)
Sum of the roots = −b/a = - (-11/1) = 11
Product of the roots = c/a = p/1 = p > 0 (Accordingly to Condition 1)

We also got sum of roots is 11, i.e. positive
We know, product of two numbers is positive if both the numbers are positive or both the numbers are negative
Since sum of two roots is 11, i.e. positive, both the numbers cannot be negative (sum of two negative numbers cannot be positive). Hence, both the roots must be positive integers.

Let the root be x1 and x2. Therefore, x1 + x2 = 11, in which both x1 and x2 are positive integers. So, we must find various ways of expressing 11 as a sum of two positive integers. Please note that zero is neither positive no negative. Hence, the first positive integer can be 1
Possible roots are (1, 10), (2, 9), (3, 8), (4, 7), and (5, 6)
p is the product of these roots, i.e. 10, 18, 24, 28, 30 = 5 values

Alternate Method:
For equation x^2 - 11x + p = 0 to have integer roots, the discriminant (D=b^2 - ac) must be greater or equal to zero AND must be a perfect square.

b^2 - ac greater or equal to zero, i.e. 4p is less than or equal to 121. Given that p > 0, so p = 1,2,3,4.......30 = 30 values
Also, b^2 - ac must be a perfect square, i.e. (121 - 4p) must be a perfect square
Between 1 and 30, only for 5 values of p, (121 - 4p) is a perfect square, 10, 18, 24, 28, 30 = 5 values

Hence, the Correct Answer is Option C. 5

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Joined: 04 Jan 2015
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If p > 0, and x2 - 11x + p = 0 has integer roots, how many integer val  [#permalink]

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01 Jan 2019, 20:41

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

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If p > 0, and x2 - 11x + p = 0 has integer roots, how many integer val   [#permalink] 01 Jan 2019, 20:41
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