smc

If the product of the roots are $7$, then by Vieta's formulas, $$2k^2-1=7$$ Solve for $k$ in the resulting equation to get $$2k^2=8$$ $$k^2=4$$ $$k=\pm 2$$ That means the two quadratics are $x^2-6x+7=0$ and $x^2+6x+7=0$. Since $b^2$, $a$, and $c$ are the same, the discriminant of both is $36-(4\cdot 1\cdot 7)=8$. Because $8$ is not a perfect square, the roots for both are irrational, so the answer is $\boxed{\text{irrational}}$.