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(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $-\alpha$ because they are equal in magnitude but opposite in sign.For

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$2$

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(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $-\alpha$ because they are equal in magnitude but opposite in sign.For any quadratic equation,the sum of the roots is given by $-\frac{b}{a}$.Here,$a = 2$,$b = 3(\lambda - 2)$,and $c = \lambda + 4$.Therefore,the sum of the roots is $\alpha + (-\alpha) = -\frac{3(\lambda - 2)}{2}$.$0 = -\frac{3(\lambda - 2)}{2}$.Multiplying both sides by $-2/3$,we get $0 = \lambda - 2$.Thus,$\lambda = 2$.

If the roots of the given equation $2x^2 + 3(\lambda - 2)x + \lambda + 4 = 0$ are equal in magnitude but opposite in sign,then $\lambda = $

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