anonymous
  • anonymous
f(x)=2x^2−146x+c has 2 roots that are positive prime numbers. What is c?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
or roots to be positive D>0 i.e. b^2 - 4ac >0 D=b^2 - 4ac = (-146)^2 - (4*2*c)>0 =>21316 - 8c>0 =>8c<21316 => c<5329/2 is dis the ans??
ParthKohli
  • ParthKohli
use the sum of roots and product of roots formulae.
ParthKohli
  • ParthKohli
\[-\dfrac{146}{2} = \alpha + \beta\]and\[\dfrac{c}{a} = \alpha \beta\]where \(\alpha\), \(\beta\) are the required prime roots.

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ParthKohli
  • ParthKohli
Know any two prime numbers with sum \(73\)?
ParthKohli
  • ParthKohli
I meant\[\dfrac{-(-146)}{2} = \alpha + \beta\]
anonymous
  • anonymous
ya its 73
anonymous
  • anonymous
but 2 prime number tat makes 73??
ParthKohli
  • ParthKohli
Yeah, do you know any two such prime numbers with sum 73?
klimenkov
  • klimenkov
If one of this primes is odd, the second must be even, because \(\alpha +\beta=73\). But how many even primes do you know?
anonymous
  • anonymous
Start from the very first prime number and work your way through. That's a hint for you.
ParthKohli
  • ParthKohli
There is only one even prime number.
anonymous
  • anonymous
I stress on the word "first" in that statement.
ParthKohli
  • ParthKohli
And of course, when you subtract the even prime, you will get another prime.
klimenkov
  • klimenkov
@ParthKohli 17 - prime. 2 - even prime. If I subtract 17 - 2 = 15 - I will get not prime.
ParthKohli
  • ParthKohli
No, in this case you must!
anonymous
  • anonymous
This case is NOT having such a situation though :-)
ParthKohli
  • ParthKohli
...?
ParthKohli
  • ParthKohli
This question would have no answer if you would not get another prime number after subtracting 2.
klimenkov
  • klimenkov
Nobody guarantees you that this must have an answer.
anonymous
  • anonymous
Hey this is a problem of the present week's Brilliant Challenges... Shouldn't be discussed here actually... Between, Awesome site for maths lovers... https://brilliant.org/
shubhamsrg
  • shubhamsrg
really? :O
anonymous
  • anonymous
Yeah...
shubhamsrg
  • shubhamsrg
woah! :O
ParthKohli
  • ParthKohli
LOL!!! ^^

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