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

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(A) Let the roots of the equation $x^2 + (k + 1)x + \lambda = 0$ be $\alpha$ and $\alpha^2$. From the properties of roots,we have: $\alpha + \alpha^2 = -(k + 1)$ $\alpha \cdot \alpha^2 = \alpha^3 = \lambda$ If the roots are squares of each other,we consider the following cases: $1$. If $\alpha = 0$,then $\alpha^2 = 0$. The equation becomes $x^2 = 0$,so $k+1 = 0 \implies k = -1$. $2$. If $\alpha = 1$,then $\alpha^2 = 1$. The equation becomes $(x-1)^2 = x^2 - 2x + 1 = 0$. Comparing with $x^2 + (k+1)x + \lambda = 0$,we get $k+1 = -2 \implies k = -3$. $3$. If $\alpha = \omega$ (where $\omega$ is a complex 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 $k+1 = 1 \implies k = 0$. 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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