If product of roots of the equation $ x^{2}-4 n x+3 e^{2 \log m}-4=0 $ is 8 , then its roots are real, when m equals

Using the product of roots formula for x^2 - 4nx + 3e^(2 log m) - 4 = 0, we set the product alpha*beta to 3e^(2 log m) - 4 = 8. Since e^(2 log m) simplifies to m^2, we solve 3m^2 - 4 = 8 to get 3m^2 = 12, meaning m equals 2 or -2. For the roots to be real, the discriminant of the equation must be non-negative, so we evaluate (-4n)^2 - 4(1)(8) which gives 16n^2 - 32 >= 0. This condition holds for any real number n except those strictly between -2 and 2, and since the question explicitly asks for the value of m, m is 2. The result is 2.

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