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    (Original post by Zacken)
    Yes, but your assumption is unjustified. Basically, you're unconciously splitting into cases:

    Case (i): assume p =/= 0 then quadratic and discriminant

    Case (ii): assume p = 0 then ....

    You can't just randomly assume p =/= 0 because you feel like and then go well "p =0 contradicts the original assumption", of course it does, you assumed p =/= 0 for no reason at all, so obviously p=0 would contradict that. You need to go back to the root and deal with the p = 0 case on its own.
    I'm not plucking the assumption out of thin air and for no reason. If we assume it is a quadratic then we are allowed to continue to inspect the roots by means of the discriminant. This would gives us the correct answers but with further inspection we would see that it cannot be p=0 otherwise it contradicts the assumption. I'm not assuming that p=/=0, I am proving this from the contradiction itself. I don't understand what you're getting at.
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    (Original post by RDKGames)
    I'm not plucking the assumption out of thin air and for no reason. If we assume it is a quadratic then we are allowed to continue to inspect the roots by means of the discriminant. This would gives us the correct answers but with further inspection we would see that it cannot be p=0 otherwise it contradicts the assumption. I'm not assuming that p=/=0, I am proving this from the contradiction itself. I don't understand what you're getting at.
    You are. What gives you the right to assume it's a quadratic? I could say "oh well, I'm gonna assume the Riemann hypothesis so I'm allowed to continue to inspect the distribution of prime number by means of the root locations, this would give us the correct answers, but with further inspection we would see...", you're arguing that p =/= 0 because it's a quadratic (but you're assuming it's a quadratic because p=/= 0, circular!). That's false. P =/= 0 because -2 =/= 0.
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    (Original post by Zacken)
    You are. What gives you the right to assume it's a quadratic? I could say "oh well, I'm gonna assume the Riemann hypothesis so I'm allowed to continue to inspect the distribution of prime number by means of the root locations, this would give us the correct answers, but with further inspection we would see...", you're arguing that p =/= 0 because it's a quadratic (but you're assuming it's a quadratic because p=/= 0, circular!). That's false. P =/= 0 because -2 =/= 0.
    Fair enough, it makes sense in my head, but I think I'm given the right to assume it's a quadratic because the equation is in the form ax^n+bx^{n-1}+...+cx+d=0 with the highest degree order of 2. Explain why this would be wrong.
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    (Original post by RDKGames)
    Fair enough, it makes sense in my head, but I think I'm given the right to assume it's a quadratic because the equation is in the form ax^n+bx^{n-1}+...+cx+d=0 with the highest degree order of 2. Explain why this would be wrong.
    Because the definition of a quadratic is a polynomial of the form ax^2 + bx + c where a \neq 0. You don't know that the coefficient of the second degree term is non-zero here, you're assuming it is.

    Anyway, this is all pedantic splitting hairs, not anything to lose much sleep over.
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    (Original post by Zacken)
    Because the definition of a quadratic is a polynomial of the form ax^2 + bx + c where a \neq 0. You don't know that the coefficient of the second degree term is non-zero here, you're assuming it is.

    Anyway, this is all pedantic splitting hairs, not anything to lose much sleep over.
    Alright, thanks for explaining.
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    (Original post by RDKGames)
    Fair enough, it makes sense in my head, but I think I'm given the right to assume it's a quadratic because the equation is in the form ax^n+bx^{n-1}+...+cx+d=0 with the highest degree order of 2. Explain why this would be wrong.
    For general polynomials you typically denote the coefficients as  a_i because there could easily be more than 26 terms in the polynomial. Even though the degree of the polynomial you have written is  n if the letters are the coefficients then it suggests that n=3. But hey, oh well.
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    (Original post by B_9710)
    For general polynomials you typically denote the coefficients as  a_i because there could easily be more than 26 terms in the polynomial. Even though the degree of the polynomial you have written is  n if the letters are the coefficients then it suggests that n=3. But hey, oh well.
    Oh well.
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    (Original post by Zacken)
    That's not the reason, nobody said it was a quadratic in the first place. The real reason why p \neq 0 is because -2 \neq 0.
    How can you tell if it is a quadratic and why would that effect the answer? Also, I don't understand why p does not equal 0 because 0 does not equal -2. Why -2?
    Thank you
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    (Original post by musicangel)
    How can you tell if it is a quadratic and why would that effect the answer? Also, I don't understand why p does not equal 0 because 0 does not equal -2. Why -2?
    Thank you
    sub p = 0 into the equation, you get 0 + 0 - 2 = 0.
 
 
 
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