If the roots of the given equation 2x^2 + 3(l | vedclass

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(B) Let the roots of the quadratic equation $2x^2 + 3(\lambda - 2)x + \lambda + 4 = 0$ be $\alpha$ and $-\alpha$.For a quadratic equation $ax^2 + bx +

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$2$

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(B) Let the roots of the quadratic equation $2x^2 + 3(\lambda - 2)x + \lambda + 4 = 0$ be $\alpha$ and $-\alpha$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of the roots is given by $-\frac{b}{a}$.Here,$a = 2$ and $b = 3(\lambda - 2)$.Sum of the roots: $\alpha + (-\alpha) = -\frac{3(\lambda - 2)}{2}$.$0 = -\frac{3(\lambda - 2)}{2}$.This implies $3(\lambda - 2) = 0$.Therefore,$\lambda - 2 = 0$,which gives $\lambda = 2$.

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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