If one root of x^2 - x - k = 0 is the square | vedclass

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(C) Let the roots of the quadratic equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the sum of roots is $\alpha + \a

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$2 \pm \sqrt{5}$

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(C) Let the roots of the quadratic equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$.From the properties of roots,the sum of roots is $\alpha + \alpha^2 = 1$ and the product of roots is $\alpha \cdot \alpha^2 = \alpha^3 = -k$.We have $\alpha^2 + \alpha - 1 = 0$. Cubing both sides of $\alpha^2 + \alpha = 1$ is not direct,so we use the identity $(\alpha^2 + \alpha)^3 = 1^3$.Expanding this,we get $(\alpha^2)^3 + \alpha^3 + 3(\alpha^2)(\alpha)(\alpha^2 + \alpha) = 1$.Substituting $\alpha^3 = -k$ and $\alpha^2 + \alpha = 1$:$(-k)^2 + (-k) + 3(-k)(1) = 1$.$k^2 - k - 3k = 1$.$k^2 - 4k - 1 = 0$.Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $k = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.

If one root of $x^2 - x - k = 0$ is the square of the other,then $k =$

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