- #1

aarciga

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## 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

ax

^{2}+ 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 [tex]\neq[/tex] 4h

so, g [tex]\neq 2h[/tex], 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 b

^{2}-4ac can't be a square number

[tex]\sqrt{4f^{2}-4(2d+1)(2(2j+1))}[/tex]

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

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

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

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

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

[tex]f^{2}-2m = \frac{S^{2}}{4} [/tex]

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.

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