Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 2207:30 PM EDT
-09:30 PM EDT
Master the GMAT with expert live instruction, a personalized study plan, and real-time support. Includes 40 hours of online classes plus 6 months of access to the TTP GMAT OnDemand video course. Class date: Mon/Wed June 22, 2026 →August 26, 2026
01 Jan 2019, 11:35
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:19) correct
42%
(02:13)
wrong
based on 283
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 6
B. 11
C. 5
D. 10
E. 0
01 Jan 2019, 13:20
Kudos
Bookmarks
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
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
General Discussion
01 Jan 2019, 20:41
Kudos
Bookmarks
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
