1. ## quadratic equation problem no.3

2. You have been given k is an >positive< integer right??
So use it

Notice that a and b are rationals and roots of x^2 + 3x + k = 0.

And if k>=3 then 9 - 4k < 0 and hence $\sqrt {9-4k}$ does not exist. This means You have to check only for k=1 and k=2, Notice that $k=1 \Rightarrow \sqrt {9-4k} = \sqrt 5$ which is irrational. But this cant hold because we have been told that roots are rationals. Thus now check for k=1........ And then claim it is the only solution.

All done

3. that means
9-4k > 0
k > 0
therefore 0<k<9/4
k=2 or 1 {rejected as x = sqrt5 [irrational] when k = 1}
thanks!