Question

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DailyChallengeQuest ​

Answer 1

Given-

• $\rm{Sum \: of \: roots ,(\alpha +\beta) }= 5$
• $\rm{Sum \: of \: cubes \: of \: root ,({\alpha}^3 +{\beta}^3) = 35}$

$\rm{where \: \alpha \: and \: \beta \: are \: roots \: of \: quadratic \: equation.}$

To Find-

• $\text{The quadratic equation}$

Solution-

$\rm{(\alpha +\beta)^3 = {\alpha}^3+{\beta}^3 + 3 \alpha \beta(\alpha +\beta)}$

Putting given values-

$\rm(5)^3 = 35 + 3 \alpha \beta(5)$

$\rm{125-35 = 15 \alpha \beta}$

$\rm{90 = 15 \alpha \beta}$

$\frac{90}{15} = \alpha \beta$

$\alpha \beta = 6$

$\boxed{ \sf{Quadratic \: Equation = x^2 + (sum \: of \: roots)x + product \: of \: roots}}$

$\implies$$\rm x^2 + (\alpha +\beta)x + \alpha \beta$

$\implies$$\large\red\star$$\boxed{\sf {x^2 + 5x + 6 }}$