a problem im having problems with...
consider the quadratic equation (-2p+1)x^2 + (p-2)x + 6p = 0
A) find the discriminant (7p-2)^2?
B) for p not equal to 1/2 show that there are always two rational solutions and find these solutions.
Could someone check if A is right and I dont know how to work out B?
thanks *

a problem im having problems with...
consider the quadratic equation (-2p+1)x^2 + (p-2)x + 6p = 0
A) find the discriminant (7p-2)^2?
B) for p not equal to 1/2 show that there are always two rational solutions and find these solutions.
Could someone check if A is right and I dont know how to work out B?
thanks *
The discriminat is

$b^2-4ac=(p-2)^2-4(-2p+1)(6p)=p^2-4p+4+48p^2-24p=49p^2-28p+4=(7p-2)^2$

correct

so by the quadratic formula the solutions are

$x=\frac{(2-p) \pm (7p-2)}{2-4p}$ this will have two distinct rational solutions unless

$7p-2=0$ if that is the case you will only have one reapeated rational solution.

$7p-2=0 \iff p=\frac{2}{7}$

I don't know where they got the one half from....