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.
To find the intersection point(s) substitute into the equation of the line , to get . 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?
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.
OK, thanks for clarifying. What you need to show is that for to be rational at least one of the coefficients must be even. Consider , where d is an integer. Rearrange: , 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.