यदि \(3x^2-11x+6=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(3x^2-11x+6=0\), what is the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?
Explanation opens after your attempt
A. \( \frac{85}{18} \)
Concept
We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha+\beta=\frac{11}{3}\) and \(\alpha\beta=2\), so the value is \(\frac{85}{18}\).
Why this answer is correct
The correct answer is A. \( \frac{85}{18} \). We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha+\beta=\frac{11}{3}\) and \(\alpha\beta=2\), so the value is \(\frac{85}{18}\).
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) होता है। यहाँ \(\alpha+\beta=\frac{11}{3}\) और \(\alpha\beta=2\), इसलिए मान \(\frac{85}{18}\) है।
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