C1 Equations

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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.

Reply 21

Zacken

Original post by RDKGames

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.

Reply 22

RDKGames

Study Forum Helper

Original post by Zacken

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.

Reply 23

Zacken

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.

(edited 7 years ago)

Reply 24

RDKGames

Study Forum Helper

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.

Reply 25

B_9710

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.

Reply 26

RDKGames

Study Forum Helper

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.

Reply 27

musicangel

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

Reply 28

Zacken

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.