If one root of the equation x^2 - x - k = 0 i | vedclass

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(A) Let the roots of the equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$. From the relation between roots and coefficients: $\alpha + \alpha^2 =

Vedclass · Accepted Answer

$2 \pm \sqrt{5}$

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(A) Let the roots of the equation $x^2 - x - k = 0$ be $\alpha$ and $\alpha^2$. From the relation between roots and coefficients: $\alpha + \alpha^2 = 1$ $\alpha \cdot \alpha^2 = -k \implies \alpha^3 = -k$ Since $\alpha^2 = 1 - \alpha$,we substitute this into $\alpha^3 = -k$: $\alpha \cdot \alpha^2 = -k \implies \alpha(1 - \alpha) = -k$ Alternatively,cube the equation $\alpha^2 + \alpha = 1$: $(\alpha^2 + \alpha)^3 = 1^3$ $\alpha^6 + \alpha^3 + 3\alpha^2 \cdot \alpha(\alpha^2 + \alpha) = 1$ Since $\alpha^3 = -k$,then $\alpha^6 = k^2$: $k^2 + (-k) + 3(-k)(1) = 1$ $k^2 - k - 3k = 1$ $k^2 - 4k - 1 = 0$ Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $k = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$

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

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