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

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(B) Given the equation $x^2 - 5x + 16 = 0$ with roots $\alpha$ and $\beta$,we have:$\alpha + \beta = 5$ and $\alpha \beta = 16$.The roots of the equat

Vedclass · Accepted Answer

$p = -1, q = -56$

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(B) Given the equation $x^2 - 5x + 16 = 0$ with roots $\alpha$ and $\beta$,we have:$\alpha + \beta = 5$ and $\alpha \beta = 16$.The roots of the equation $x^2 + px + q = 0$ are $\alpha^2 + \beta^2$ and $\frac{\alpha \beta}{2}$.First,calculate the sum of the roots:$(\alpha^2 + \beta^2) + \frac{\alpha \beta}{2} = -p$Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = 5^2 - 2(16) = 25 - 32 = -7$,$-7 + \frac{16}{2} = -p \Rightarrow -7 + 8 = -p \Rightarrow 1 = -p \Rightarrow p = -1$.Next,calculate the product of the roots:$(\alpha^2 + \beta^2) \cdot \frac{\alpha \beta}{2} = q$$(-7) \cdot \frac{16}{2} = q \Rightarrow -7 \cdot 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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