Question

Show that if `a, b, and c` are all odd integers then the quadratic `ax^2 + bx + c = 0` has no solution in the set of rational numbers?

Answer 1

The discriminant of a quadratic formula is `b^2-4ac`.

For a number to be odd, `o=2n+1, ninNN`

Using this, we get: `(2b+1)^2-4(2a+1)(2c+1)`

By expanding, we get: `4b^2+b-16ac-8c-8a+5`. No matter what integers you use, you get an irrational number when the quare root is taken.

Proof:
`a=3,b=7,c=-1`
`4(7)^2+4(7)-16(3*-1)-8(-1)-8(3)+5=261`
`sqrt(261)~~16.2`

`a=-5,b=3,c=9`
`4(3)^2+4(3)-16(-5*9)-8(9)-8(-5)+5=741`
`sqrt(261)~~27.2`

Also, `(2n+1)^2+(2n+3)^2=4n^2+4n+1+4n^2+12n+9=8n^2+16n+10`, this would always give an irrational number whenever the answer is square rooted. This can be easily proved, to do this, I will use `3^2+5^2`. I will be using matrices to help represent a square grid.

`[(1,1,1),(1,1,1),(1,1,1)]+[(1,1,1,1,1),(1,1,1,1,1),(1,1,1,1,1),(1,1,1,1,1),(1,1,1,1,1)]`
`=[(1,1,1,1,1,1),(1,1,1,1,1,1),(1,1,1,1,1,1),(1,1,1,1,1,1),(1,1,1,1,1,1),(1,1,1,1,0,0)]`

As you can see, there are two spaces missing.

Though, if the squares of two odd numbers are multiplied together, it still gives a square number, `a^2b^2=(ab)^2`

`(2b+1)^2-4(2a+1)(2c+1)` will give us two new square numbers, but based on the proof with `3^2` and `5^2`, it will give an incomplete square, and therefore an irrational number.

Also, if we take `(2n+1)^2+(2n+3)^2=m^2`

`m^2=8n^2+16n+10`

`8n^2+16n+(10-m^2)=0`

`n=(-16+-sqrt(16^2-4(80-8m^2)))/16`

`n=(-16+-sqrt(-64+32m^2))/16`

No matter what integer you use for `m` as long as it is greater than `sqrt(2)`, `n` will never be a rational number. For `n` to be a rational number, `m=sqrt((64^d+64)/2), dinZZ, d>=2` but `m` will be an irrational number.

Answer 2

See explanation...

Explanation:

If `a, b, c` are odd integers, then they are all non-zero.

If `ax^2+bx+c` has a rational zero, then it can be factored into two linear factors with integer coefficients, say:

`ax^2+bx+c = (px+q)(rx+s)`

`color(white)(ax^2+bx+c) = prx^2+(ps+qr)x+qs`

Equating coefficients, we have:

`{ (a = pr), (b=ps+qr), (c=qs) :}`

Since `a` and `c` are both odd integers, so must be `p, r, q, s`.

Then:

`b = ps+qr`

is the sum of two odd integers and therefore even.

This contradicts the assumption that `b` is odd. So there is no such factorisation into linear terms with integer coefficients. Therefore `ax^2+bx+c` has no rational zeros.