If the roots of the equation x^2 - 5x + 16 = | vedclass

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Question

(B) Given the roots of $x^2 - 5x + 16 = 0$ are $\alpha, \beta$,we have $\alpha + \beta = 5$ and $\alpha \beta = 16$.For the equation $x^2 + px + q = 0

Vedclass · Accepted Answer

$p = -1, q = -56$

Vedclass · Answer

(B) Given the roots of $x^2 - 5x + 16 = 0$ are $\alpha, \beta$,we have $\alpha + \beta = 5$ and $\alpha \beta = 16$.For the equation $x^2 + px + q = 0$,the roots are $\alpha^2 + \beta^2$ and $\frac{\alpha \beta}{2}$.Sum of roots: $(\alpha^2 + \beta^2) + \frac{\alpha \beta}{2} = -p$.Using $(\alpha + \beta)^2 - 2\alpha \beta = \alpha^2 + \beta^2$,we get $(5^2 - 2(16)) + \frac{16}{2} = -p$.$(25 - 32) + 8 = -p$ $\Rightarrow -7 + 8 = -p$ $\Rightarrow 1 = -p$ $\Rightarrow p = -1$.Product of roots: $(\alpha^2 + \beta^2) 	imes \frac{\alpha \beta}{2} = q$.$(25 - 32) 	imes \frac{16}{2} = q$ $\Rightarrow (-7) 	imes 8 = q$ $\Rightarrow q = -56$.Thus,$p = -1$ and $q = -56$.

If the roots of the equation $x^2 - 5x + 16 = 0$ are $\alpha, \beta$ and the roots of the equation $x^2 + px + q = 0$ are $\alpha^2 + \beta^2$ and $\frac{\alpha \beta}{2}$,then:

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