If $x^2-2013x+c=0$ and the both roots of this equation is prime numbers, what is the possible number for $^3√c$?
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Hint:
So, the sum of prime roots is $=\dfrac{2013}1$
Now if both roots are odd, the sum will be even
lab bhattacharjee
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1Oh i get it then x1= 2 as it is even X2= 2011, and c = 4022 okay, thank you so much for your help – ilham maha Jun 05 '20 at 15:35
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Give the man a gold star! – Oscar Lanzi Jun 05 '20 at 15:38
1
Hint:
You're looking for two prime numbers (the roots) adding up to $2013$. That sum is odd and so the individual roots must be one of them odd and the other even.
I guess we'll have to use brute force here, try out all the even primes we can think of for the even root, work out the odd root for each case and check that this odd number is also prime.
Oscar Lanzi
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