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arun8408

f(x)=2x^2−146x+c has 2 roots that are positive prime numbers. What is c?

  • one year ago
  • one year ago

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  1. arun8408
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    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??

    • one year ago
  2. ParthKohli
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    use the sum of roots and product of roots formulae.

    • one year ago
  3. ParthKohli
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    \[-\dfrac{146}{2} = \alpha + \beta\]and\[\dfrac{c}{a} = \alpha \beta\]where \(\alpha\), \(\beta\) are the required prime roots.

    • one year ago
  4. ParthKohli
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    Know any two prime numbers with sum \(73\)?

    • one year ago
  5. ParthKohli
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    I meant\[\dfrac{-(-146)}{2} = \alpha + \beta\]

    • one year ago
  6. arun8408
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    ya its 73

    • one year ago
  7. arun8408
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    but 2 prime number tat makes 73??

    • one year ago
  8. ParthKohli
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    Yeah, do you know any two such prime numbers with sum 73?

    • one year ago
  9. klimenkov
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    If one of this primes is odd, the second must be even, because \(\alpha +\beta=73\). But how many even primes do you know?

    • one year ago
  10. Azteck
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    Start from the very first prime number and work your way through. That's a hint for you.

    • one year ago
  11. ParthKohli
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    There is only one even prime number.

    • one year ago
  12. Azteck
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    I stress on the word "first" in that statement.

    • one year ago
  13. ParthKohli
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    And of course, when you subtract the even prime, you will get another prime.

    • one year ago
  14. klimenkov
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    @ParthKohli 17 - prime. 2 - even prime. If I subtract 17 - 2 = 15 - I will get not prime.

    • one year ago
  15. ParthKohli
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    No, in this case you must!

    • one year ago
  16. saloniiigupta95
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    This case is NOT having such a situation though :-)

    • one year ago
  17. ParthKohli
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    ...?

    • one year ago
  18. ParthKohli
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    This question would have no answer if you would not get another prime number after subtracting 2.

    • one year ago
  19. klimenkov
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    Nobody guarantees you that this must have an answer.

    • one year ago
  20. saloniiigupta95
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    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/

    • one year ago
  21. shubhamsrg
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    really? :O

    • one year ago
  22. saloniiigupta95
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    Yeah...

    • one year ago
  23. shubhamsrg
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    woah! :O

    • one year ago
  24. ParthKohli
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    LOL!!! ^^

    • one year ago
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