The values of p for which one root of the equ | vedclass

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(B) Let the roots of the equation $x^2 - 30x + p = 0$ be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients,the sum of roots is

Vedclass · Accepted Answer

$125$ and $-216$

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(B) Let the roots of the equation $x^2 - 30x + p = 0$ be $\alpha$ and $\alpha^2$.From the relation between roots and coefficients,the sum of roots is $\alpha + \alpha^2 = 30$ and the product of roots is $\alpha \cdot \alpha^2 = \alpha^3 = p$.Solving $\alpha^2 + \alpha - 30 = 0$:$(\alpha + 6)(\alpha - 5) = 0$,which gives $\alpha = -6$ or $\alpha = 5$.If $\alpha = 5$,then $p = \alpha^3 = 5^3 = 125$.If $\alpha = -6$,then $p = \alpha^3 = (-6)^3 = -216$.Thus,the values of $p$ are $125$ and $-216$.

The values of $p$ for which one root of the equation $x^2 - 30x + p = 0$ is the square of the other are:

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