# Thread: straight line and function

1. ## straight 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.

2. ## Re: straight line and function

To find the intersection point(s) substitute $y = x^2$ into the equation of the line $ax+by+c = 0$, to get $ax + bx^2+c = 0$. 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?

3. ## Re: straight line and function

thank you ebaines, but it still not clear to me

4. ## Re: straight line and function

Originally Posted by abualabed
thank you ebaines, but it still not clear to me
Do you know the quadratic formula for finding the roots of a quadratic equation?

5. ## Re: straight line and function

Shameful to ask me this question , ebaines

6. ## Re: straight line and function

Originally Posted by abualabed
Shameful to ask me this question , ebaines

What are the roots of $bx^2+ax+c=0~?$

ok

8. ## Re: straight line and function

Originally Posted by abualabed
ok
If $K\in\mathbb{Z}^+$, then do you know under what conditions $\sqrt{K}$ is rational?
What does that have to do with the OP?

9. ## Re: straight line and function

Originally Posted by abualabed
ok
So if $\sqrt {a^2-4bc} = 0$ there would be only one root for $bx^2+ax+c=0$ 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.

10. ## Re: straight line and function

Originally Posted by ebaines
So if $\sqrt {a^2-4bc} = 0$ there would be only one root for $bx^2+ax+c=0$ 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.
@ebaines, Frankly, I don't understand your point.
If $a=3,~b=1,~\&~c=1$ the line intersects in two point.
But neither of the points has a rational x-coordinate. Why?

Have a look at this,

11. ## Re: straight line and function

Originally Posted by Plato
@ebaines, Frankly, I don't understand your point.
If $a=3,~b=1,~\&~c=1$ the line intersects in two point.
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.

12. ## Re: straight line and function

Originally Posted by ebaines
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.
I could see reading it that way if the word tangent had been used. Otherwise, I disagree with that reading.

I realize that this is a translated question. So who really knows.

13. ## Re: straight line and function

hello Plato, ebaines

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

14. ## Re: straight line and function

Originally Posted by abualabed
hello Plato, ebaines
I mean that" the x-coordinate of any intersection point never be rational number"
That is exactly how I read the question and my replies reflect that reading.

15. ## Re: straight line and function

OK, thanks for clarifying. What you need to show is that for $\sqrt{a^2-4bc}$ to be rational at least one of the coefficients must be even. Consider $a^2-4bc = d^2$, where d is an integer. Rearrange: $a^2-d^2 = 4bc$, 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.

Page 1 of 2 12 Last