If the roots of the quadratic equation $2kx^{2}+(4k-1)x+2k-3=0$ are rational and k is an integer, how many values can k take which are less that 50 ?
The discriminant = $16k+1$
For a rational number the discriminant = $p^{2}$
i.e. $16k=p^{2}-1$
$16k=(p-1)(p+1)$
$k=\frac{(p-1)(p+1)}{16}$
$16=16*1=8*2=4*4=2*8=1*16$
So far so good. Notice that the equation $16k=p^2-1$ shows that $p^2$ is congruent to $1$ modulo $16$. You can then show that $p$ has to be congruent to $1,7,9,$ or $15$ modulo $16$. This gives a list of possible $p$'s: $1,7,9,15,17,23,25,31,\ldots$. You'll find that $p=31$ gives $k=60$, but everything smaller on the list should work. It follows that there are $7$ such values of $k$ (if you include $k=0$).
Yes, but you should also consider multiples of $16$. For example, $p^2-1=48$ has solutions $p=+7$ and $p=-7$, so in your case $16*k=48$ and $k=3$. This is just one solution.
