Prove by contradiction that if a and b are integers and b is odd, then -1 is not a root of f(x)=ax^2+bx+c

Question

anonymous · Answer

I mean ax^2+bx+a

anonymous · Answer

0 = a - b + c
b = a + c

nothing about a or c ?

myininaya · Answer

Ok b is odd=>b=2k+1, k is an element of an integer
ok so we have
$$f(x)=ax^2+(2k+1)x+a$$
so its's possible for (-1)=0
well lets see..
$$f(-1)=a(-1)^2+(2k+1)(-1)+a=a-(2k+1)+a$$
$$=2a-(2k+1)$$
is it possible for 2a-(2k+1)=0 well if a =(2k+1)/2
but this means a is not an integer
thus we have received a contradiction

myininaya · Answer

no odd number is divisible by 2

myininaya · Answer

so its's possible for f(-1)=0*

anonymous · Answer

why c = a ?

myininaya · Answer

thats what he said

anonymous · Answer

Thank you SO much. Are you a math type of major?

anonymous · Answer

lol i havent looked on the second thing he wrote :\

anonymous · Answer

0 = a - b + a
b = 2a
but 2a is even
so b must be even as well