If both roots of the equation x^2 + lambda x | vedclass

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Question

(A) Since the roots of the equation $x^2 + \lambda x + \mu = 0$ are equal,the discriminant $D = 0$. Thus,$\lambda^2 - 4\mu = 0 \implies \lambda^2 = 4\

Vedclass · Accepted Answer

$(4, 4)$

Vedclass · Answer

(A) Since the roots of the equation $x^2 + \lambda x + \mu = 0$ are equal,the discriminant $D = 0$. Thus,$\lambda^2 - 4\mu = 0 \implies \lambda^2 = 4\mu$. Given that $x = 2$ is a root of the equation $x^2 + \lambda x - 12 = 0$,we substitute $x = 2$ into the equation: $2^2 + \lambda(2) - 12 = 0 4 + 2\lambda - 12 = 0 2\lambda = 8 \lambda = 4$. Now,substitute $\lambda = 4$ into $\lambda^2 = 4\mu$: $4^2 = 4\mu 16 = 4\mu \mu = 4$. Therefore,$(\lambda, \mu) = (4, 4)$.

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

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