Both roots of the quadratic equation x^2-39x+a = 0 are prime

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[GMAT math practice question]

Both roots of the quadratic equation x^2-39x+a = 0 are prime numbers. How many different possible values of a are there?

A.1
B. 2
C. 3
D. 4
E. 5
Last edited by Max@Math Revolution on Sun Jun 30, 2019 4:51 pm, edited 1 time in total.

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by deloitte247 » Sun Jun 30, 2019 2:33 pm
$$x^{^2}+39x+9=0$$
sum of roots = a + b =39
Products of roots = ab = 1a
Given that the sum of roots is odd number that means
a +b = Odd + Even = Odd number
Both a and b are prime number and one of them is even. The only prime number that is an even number is 2
hence
a + b = 39
Odd + Even = 39
Odd + 2 = 39
a = 39 - 2 = 37
a = 37 and b = 2
ab = 37*2 =74
There is only one possible value of measuring variable. a

$$answer\ is\ Option\ A.$$

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by Max@Math Revolution » Sun Jun 30, 2019 4:51 pm
=>

Assume p and q are roots of x^2-39x+a = 0.
Then (x-p)(x-q) = x^2 -(p+q)x + pq = x^2-39x+a = 0
So, p + q = 39 and pq = a.
Since p and q are prime numbers and p + q = 39 is an odd number, one of p and q is an even prime number. Since the only even prime number is 2, one of p and q must be 2.
Let p = 2. Then q = 37,
and a = 2*37 = 74.
There is only one possible value of a.

Therefore, the answer is A.
Answer: A