यदि \(x^2-6x+3=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+3=0\), what is \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\)?
Explanation opens after your attempt
Correct Answer
B. \(\frac{10}{3}\)
Concept
We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=30\) and (\(\alpha\beta\)2=9), the value is \(\frac{10}{3}\).
Why this answer is correct
The correct answer is B. \(\frac{10}{3}\). We use (\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}). Since \(\alpha^2+\beta^2=30\) and (\(\alpha\beta\)2=9), the value is \(\frac{10}{3}\).
Exam Tip
(\frac{1}{\alpha-2}+\frac{1}{\beta-2}=\frac{\alpha-2+\beta-2}{\(\alpha\beta\)2}) होता है। \(\alpha^2+\beta^2=30\) और (\(\alpha\beta\)2=9), इसलिए मान \(\frac{10}{3}\) है।
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