If one root of the equation $x^2-\lambda x+12=0$ is even prime while $x^2 +\lambda x+\mu =0$ has equal roots, then $\mu$ is equal to

The only even prime number is 2, so substituting x=2 into the first equation x^2 - lx + 12 = 0 gives 4 - 2l + 12 = 0. Solving for l yields 2l = 16, so l = 8. The second equation x^2 + lx + u = 0 has equal roots, meaning its discriminant must be zero: l^2 - 4u = 0. Substituting l = 8 gives 64 - 4u = 0, which results in u = 16.

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