If alpha, beta are the roots of the equation | vedclass

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Question

(D) Given the equation: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\lef

Vedclass · Accepted Answer

$3x^{2}-10x-4=0$

Vedclass · Answer

(D) Given the equation: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$. Simplifying the coefficients: $3^{\sqrt{\log _{3} 5}} = 3^{\sqrt{\log _{3} 5} \cdot \frac{\sqrt{\log _{3} 5}}{\sqrt{\log _{3} 5}}} = 3^{\log _{3} 5 \cdot \sqrt{\log _{5} 3}} = (3^{\log _{3} 5})^{\sqrt{\log _{5} 3}} = 5^{\sqrt{\log _{5} 3}}$. Thus,the coefficient of $x$ is $-(5 + 5^{\sqrt{\log _{5} 3}} - 5^{\sqrt{\log _{5} 3}}) = -5$. Similarly,$3^{\left(\log _{3} 5\right)^{\frac{1}{3}}} = 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}$. Thus,the constant term is $3(5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} - 1) = -3$. The equation simplifies to $x^{2}-5x-3=0$. For this equation,$\alpha+\beta=5$ and $\alpha\beta=-3$. The new roots are $\alpha+\frac{1}{\beta} = \frac{\alpha\beta+1}{\beta} = \frac{-3+1}{\beta} = \frac{-2}{\beta}$ and $\beta+\frac{1}{\alpha} = \frac{-2}{\alpha}$. Let $t = \frac{-2}{\alpha}$,then $\alpha = \frac{-2}{t}$. Substituting into $x^{2}-5x-3=0$: $(\frac{-2}{t})^{2} - 5(\frac{-2}{t}) - 3 = 0 \Rightarrow \frac{4}{t^{2}} + \frac{10}{t} - 3 = 0$. Multiplying by $t^{2}$: $4 + 10t - 3t^{2} = 0 \Rightarrow 3t^{2}-10t-4=0$. Therefore,the required equation is $3x^{2}-10x-4=0$.

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