यदि \(x^2-11x+24=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-11x+24=0\), what is \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)?
Explanation opens after your attempt
Correct Answer
C. \(\frac{73}{24}\)
Concept
We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=73\) and \(\alpha\beta=24\), so the value is \(\frac{73}{24}\).
Why this answer is correct
The correct answer is C. \(\frac{73}{24}\). We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=73\) and \(\alpha\beta=24\), so the value is \(\frac{73}{24}\).
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) है। यहाँ \(\alpha^2+\beta^2=73\) और \(\alpha\beta=24\), इसलिए मान \(\frac{73}{24}\) है।
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