quadratic equation problem no.3
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 $\displaystyle \sqrt {9-4k}$ does not exist. This means You have to check only for k=1 and k=2, Notice that $\displaystyle k=1 \Rightarrow \sqrt {9-4k} = \sqrt 5$ 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