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Joined: 29 Nov 2018
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Concentration: Marketing, Strategy

If p > 0, and x2  11x + p = 0 has integer roots, how many integer val
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01 Jan 2019, 11:35
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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 If you liked the question, please do hit the kudos button
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Re: If p > 0, and x2  11x + p = 0 has integer roots, how many integer val
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01 Jan 2019, 13:20
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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If p > 0, and x2  11x + p = 0 has integer roots, how many integer val
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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 Answer: C
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If p > 0, and x2  11x + p = 0 has integer roots, how many integer val
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01 Jan 2019, 20:41






