Because b^2-4ac > 0 and also a perfect square, the roots are real and rational.

Given information
• L M N = 0
• L M N are rationals

To find
• The nature of roots of the equation (M NL) x2 (N LM)x (L MN) = 0

Solution
Let's assume α and β be the roots of the given equation.

Sum of roots = - (N LM)/ (M NL) = -N/M
Product of roots = (L MN)/ (M NL) = N/L

Using these values, we can write the quadratic equation as:
x2 - (sum of roots) x + (product of roots) = 0
x2 + (N/M)x + (N/L) = 0

Now, we need to check the nature of roots of the quadratic equation.

Discriminant = b2 - 4ac
= (N/M)2 - 4(N/L)(L/M)
= (N2/M2) - 4(N2/LM)
= N2(1/M2 - 4/LM)

As we know that L M N = 0, which implies that at least one of the value should be zero.
Let's consider the possibilities:
• If L = 0, then the equation will become x2 = 0, which will have only one real and rational root, i.e., x = 0. But in the given equation, L MN ≠ 0, which means L cannot be zero.
• Similarly, we can prove that M and N cannot be zero.

Hence, we can assume that L ≠ 0, M ≠ 0, and N ≠ 0.

Now, coming back to the discriminant:
N2(1/M2 - 4/LM) = N2(L - 4M)/M2L

Since L and M are rationals, their difference (L - 4M) will also be a rational number. Hence, the discriminant will be positive if and only if N2/M2L > 0. This implies that the signs of N and L should be the same.

Therefore, the possible cases are:
• If L, M, and N are all positive or all negative, then the discriminant will be positive, and the roots will be real and rational.
• If L and M are positive/negative, and N is negative/positive, then the discriminant will be negative, and the roots will be imaginary and unequal.

Hence, the correct option is (B) real and rational.

Conclusion
The nature of roots of the given equation (M NL) x2 (N LM)x (L MN) = 0 is real and rational.