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(B) Given the quadratic equation $x^2 + 6x + \lambda = 0$,the sum of the roots is $\alpha + \beta = -6$ $(i)$ and the product of the roots is $\alpha

Vedclass · Accepted Answer

$-16$

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(B) Given the quadratic equation $x^2 + 6x + \lambda = 0$,the sum of the roots is $\alpha + \beta = -6$ $(i)$ and the product of the roots is $\alpha \beta = \lambda$ $(ii)$.We are also given the equation $3\alpha + 2\beta = -20$ $(iii)$.From $(i)$,we have $\beta = -6 - \alpha$. Substituting this into $(iii)$:$3\alpha + 2(-6 - \alpha) = -20$$3\alpha - 12 - 2\alpha = -20$$\alpha = -20 + 12 = -8$.Now,find $\beta$ using $(i)$:$\beta = -6 - (-8) = 2$.Finally,substitute $\alpha$ and $\beta$ into $(ii)$:$\lambda = \alpha \beta = (-8)(2) = -16$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 6x + \lambda = 0$ and $3\alpha + 2\beta = -20$,then $\lambda = $

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