If the equation x^2 + lambda x + mu = 0 has e | vedclass

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(A) For the equation $x^2 + \lambda x + \mu = 0$ to have equal roots,the discriminant must be zero,i.e.,$D = \lambda^2 - 4\mu = 0$,which implies $\lam

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$(4, 4)$

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(A) For the equation $x^2 + \lambda x + \mu = 0$ to have equal roots,the discriminant must be zero,i.e.,$D = \lambda^2 - 4\mu = 0$,which implies $\lambda^2 = 4\mu$.Given that $x = 2$ is a root of the equation $x^2 + \lambda x - 12 = 0$,we substitute $x = 2$ into the equation:$2^2 + \lambda(2) - 12 = 0$$4 + 2\lambda - 12 = 0$$2\lambda - 8 = 0$$2\lambda = 8$$\lambda = 4$Now,substitute $\lambda = 4$ into the relation $\lambda^2 = 4\mu$:$4^2 = 4\mu$$16 = 4\mu$$\mu = 4$Therefore,$(\lambda, \mu) = (4, 4)$.

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $

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