The quadratic equation whose sum of the roots | vedclass

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(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation. Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$. We know that $(\alpha+\beta)^2

Vedclass · Accepted Answer

$x^2-11x+30=0$

Vedclass · Answer

(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation. Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$. We know that $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta$. Substituting the values: $(11)^2 = 61 + 2\alpha\beta$. $121 = 61 + 2\alpha\beta$ $\Rightarrow 2\alpha\beta = 60$ $\Rightarrow \alpha\beta = 30$. The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Substituting the values: $x^2 - 11x + 30 = 0$.

The quadratic equation whose sum of the roots is $11$ and sum of squares of the roots is $61$ is

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