if a , b , c are odd numbers prove that the straight line ax+by+c=0 never

intersect the function f(x)=x^2 in apoint its x-coordinate is rational number.

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- May 7th 2013, 02:14 AMabualabedstraight line and function
**if a , b , c are odd numbers prove that the straight line ax+by+c=0 never**

intersect the function f(x)=x^2 in apoint its x-coordinate is rational number.

- May 7th 2013, 05:05 AMebainesRe: straight line and function
To find the intersection point(s) substitute $\displaystyle y = x^2$ into the equation of the line $\displaystyle ax+by+c = 0$, to get $\displaystyle ax + bx^2+c = 0$. Now apply the quadratic formula to solve for values of x. Recall that in general the quadratic equation gives two roots (due to the "plus or minus" in front of the square root), unless the value of the square root is zero. So - given that the values a, b, and c are all odd, can that square root = 0?

- May 7th 2013, 05:38 AMabualabedRe: straight line and function
thank you ebaines, but it still not clear to me

- May 7th 2013, 05:45 AMebainesRe: straight line and function
- May 7th 2013, 06:57 AMabualabedRe: straight line and function
Shameful to ask me this question , ebaines

- May 7th 2013, 07:00 AMPlatoRe: straight line and function
- May 7th 2013, 07:06 AMabualabedRe: straight line and function
ok

- May 7th 2013, 07:24 AMPlatoRe: straight line and function
- May 7th 2013, 08:24 AMebainesRe: straight line and function
So if $\displaystyle \sqrt {a^2-4bc} = 0$ there would be only one root for $\displaystyle bx^2+ax+c=0$ right? But if a, b, and c are all odd, is this possible? If not, then there must be either two real or two complex intersection points between the line and the parabola, not just one.

- May 7th 2013, 08:51 AMPlatoRe: straight line and function
@ebaines, Frankly, I don't understand your point.

If $\displaystyle a=3,~b=1,~\&~c=1$ the line intersects in two point.

But neither of the points has a rational-coordinate. Why?*x*

Have a look at this, - May 7th 2013, 09:21 AMebainesRe: straight line and function
I interpret the OP's problem statement as proving that if a, b, and c are all odd then the line and parabola cannot touch at one point only. His phrase was: "straight line ax+by+c=0 never intersect the function f(x)=x^2 in apoint " I believe by "intersect ... in a point" he means the two functions touching at one point only, not two.

Example: if a=6, b=3 and c=3 the two lines meet at only one point, at (-1,1). This is possible because a, b. and c are not all odd. But if a= 5, b=3, c = 1 they intersect at two points, and if a=5, b=3, c=3 they don't intersect at all. - May 7th 2013, 09:33 AMPlatoRe: straight line and function
- May 7th 2013, 10:34 AMabualabedRe: straight line and function
hello Plato, ebaines

I mean that"__the x-coordinate of any intersection point never be rational number"__

- May 7th 2013, 10:38 AMPlatoRe: straight line and function
- May 7th 2013, 10:48 AMebainesRe: straight line and function
OK, thanks for clarifying. What you need to show is that for $\displaystyle \sqrt{a^2-4bc}$ to be rational at least one of the coefficients must be even. Consider $\displaystyle a^2-4bc = d^2$, where d is an integer. Rearrange: $\displaystyle a^2-d^2 = 4bc$, so if 'a' is odd then 'd' must also be odd since the right hand side is even. It's not too difficult to show that the difference of two odd numbers squared must be a mutiple of 8; therefore either 'b' or 'c' must be even.