$(\text{Solve the following equations:})\newline(\text{(i) } 2x^2 + 31x + 119 = 0\newline\text{(ii) } y^2 + 51y + 98 = 0)$

For equation (i): 2x^2 + 31x + 119 = 0. Using quadratic formula: x = [-31 ± √(961 - 952)]/4 = [-31 ± √9]/4 = [-31 ± 3]/4. Roots are x = -34/4 = -8.5 and x = -28/4 = -7. For equation (ii): y^2 + 51y + 98 = 0. Roots are y = [-51 ± √(2601 - 392)]/2 = [-51 ± √2209]/2 = [-51 ± 47]/2. Roots are y = -98/2 = -49 and y = -4/2 = -2. So x can be -7 or -8.5, while y can be -2 or -49. Sometimes x > y (when x=-7 and y=-49), sometimes x < y (when x=-8.5 and y=-2). No definite relationship. Option E is correct.