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(A) Let the roots of the quadratic equation $x^2 + (k + 1)x + \lambda = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the product of the

Vedclass · Accepted Answer

$-4$

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(A) Let the roots of the quadratic equation $x^2 + (k + 1)x + \lambda = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the product of the roots is $\alpha \cdot \alpha^2 = \alpha^3 = \lambda$.The sum of the roots is $\alpha + \alpha^2 = -(k + 1)$.Cubing the sum equation: $(\alpha + \alpha^2)^3 = (-(k + 1))^3$.$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -(k + 1)^3$.Substituting $\alpha^3 = \lambda$ and $\alpha + \alpha^2 = -(k + 1)$:$\lambda + \lambda^2 + 3\lambda(-(k + 1)) = -(k + 1)^3$.$\lambda^2 + \lambda - 3\lambda(k + 1) = -(k + 1)^3$.For the roots to be squares of each other,we consider specific cases:$1$. If $\alpha = 0$,then $\lambda = 0$. The equation becomes $x^2 + (k + 1)x = 0$. Roots are $0, -(k+1)$. For $0^2 = -(k+1)$,$k = -1$.$2$. If $\alpha = 1$,then $\lambda = 1$. The equation becomes $x^2 + (k + 1)x + 1 = 0$. Roots are $1, 1$. Sum $1+1 = -(k+1) \implies k = -3$.$3$. If $\alpha = \omega$ (cube root of unity),then $\alpha^2 = \omega^2$. The equation is $x^2 + x + 1 = 0$. Comparing with $x^2 + (k+1)x + \lambda = 0$,we get $k+1 = 1 \implies k = 0$ and $\lambda = 1$.The possible values of $k$ are $-1, -3, 0$.The sum of these values is $(-1) + (-3) + 0 = -4$.

Find the sum of all possible values of $k$ for which the roots of the equation $x^2 + (k + 1)x + \lambda = 0$ are the square of each other.

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