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15 Feb 2022, 23:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
55% (02:17) correct
45%
(02:03)
wrong
based on 148
sessions
History
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Time
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Not Attempted Yet
Both roots of the quadratic equation \(x^2 - 63*x + k = 0\) are prime numbers. The number of possible values of k are?
A. 0
B. 1
C. 2
D. 3
E. 4 or more
Source: Chinese GMAT practice tests book
A. 0
B. 1
C. 2
D. 3
E. 4 or more
Source: Chinese GMAT practice tests book
16 Feb 2022, 02:30
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In the standard form of a quadratic equation, \(ax^2 + bx + c = \), we know that c will be the product of the two factors, while b will be the sum of the two factors.
We are told that the roots of the quadratic are both prime numbers, and the only way one can have \(bx\) be an odd number with two primes is if one of the roots is 2 (odd + odd = even, even + odd = odd), which means that the other root is 61.
Therefore there is only 1 possible value for k.
Answer B
We are told that the roots of the quadratic are both prime numbers, and the only way one can have \(bx\) be an odd number with two primes is if one of the roots is 2 (odd + odd = even, even + odd = odd), which means that the other root is 61.
Therefore there is only 1 possible value for k.
Answer B
19 Apr 2023, 01:24
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twobagels
Solution:
- Let the roots of \(x^2 - 63x + k = 0\) be \(P_1\) and \(P_2\)
- So, we can infer \(P_1+P_2=63\)
- Sum of two primes will be 63 (odd) iff one of them is 2 (even)
- So, \(P_1=2\) and \(P_2=61\) are the only roots possible
- For one pair of x, we will have only one value of k
Hence the right answer is Option B
