If alpha and beta are the roots of the equati | vedclass

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(A) Given $\alpha+\beta=a$ and $\alpha\beta=b$. We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2-2b$. Also,$\alpha^3+\beta^3 = (

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$a^5-5a^3b+6ab^2$

Vedclass · Answer

(A) Given $\alpha+\beta=a$ and $\alpha\beta=b$. We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2-2b$. Also,$\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) = a(a^2-2b-b) = a(a^2-3b) = a^3-3ab$. The quadratic equation with roots $p$ and $q$ is given by $x^2 - (p+q)x + pq = 0$. Here,the roots are $p = a^2-2b$ and $q = a^3-3ab$. The constant term $C$ is the product of the roots: $C = pq = (a^2-2b)(a^3-3ab)$. Expanding this: $C = a^2(a^3) - a^2(3ab) - 2b(a^3) + 2b(3ab) = a^5 - 3a^3b - 2a^3b + 6ab^2 = a^5 - 5a^3b + 6ab^2$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2-ax+b=0$,and $\alpha^2+\beta^2$ and $\alpha^3+\beta^3$ are the roots of the equation $Ax^2+Bx+C=0$,then $C=$

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