Proof of quadratic irrationals - stuck

[aarciga]

Homework Statement

I have to show a proof that if a is odd, b is even, and c is even but not divisible by 4.
a,b,c are int coefficients

ax² + bx + c = 0

has no rational solutions

Homework Equations

all letters here are integers.

So i have,

a = 2d + 1
b = 2f
c = 2g, but c is not divisable by for so, c \neq 4h
so, g \neq 2h, or g is not even. g is odd
c = 2(2j + 1)

The Attempt at a Solution

Im trying to show that the discriminant in the quadratic formula can't be a square, but I am having trouble showing that.

or that b²-4ac can't be a square number

\sqrt{4f^{2}-4(2d+1)(2(2j+1))}

(2d+1)(2j+1) will give some odd number 2k+1

\sqrt{4[f^{2}-2(2k+1)]}

2(2k+1) will give an even number 2m

\sqrt{4[f^{2}-2m]}

\sqrt{f^{2}-2m}^{2} can't = some square \frac{S^{2}}{2^{2}}

f^{2}-2m = \frac{S^{2}}{4}

After I show that it isn't a square, i can show that the square root of a non square is irrational. But I feel like i might be off base on this. It seems like the discrimininant certainly could be a perfect square.

it looks like I am trying to show that a perfect square can't equal a perfect square divided by 4. which is false. ( i.e. 16/4 = 4)


so, any ideas to point me in more proper direction, or to make a point on what i already have would be greatly appreciated.

 

[aarciga]

Another way of thinking about it??


\sqrt{b^{2}-4ac}

\sqrt{(2f)^{2}-4(2d+1)(2(2j+1))}

2f is even, so (2f)² is even
and 8(2d+1)(2j+1) is even x (odd x odd). (odd x odd) is odd so then even x odd = even

\sqrt{even number -even number}
even number - even number = even. = 2m

so i must show
\sqrt{2m} can't be rational, or

\sqrt{2m} \neq some rational number \frac{P}{Q}


and from here, I can prove \sqrt{2} is irrational.

and assuming \sqrt{m} is rational, an irrational time a rational can't be rational??

\sqrt{2}\sqrt{m} \neq \frac{P}{Q}