यदि \(x^2-10x+q=0\) के मूल \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=58\) है, तो (q) का मान क्या होगा?
If \(\alpha,\beta\) are roots of \(x^2-10x+q=0\) and \(\alpha^2+\beta^2=58\), what is the value of (q)?
Explanation opens after your attempt
Correct Answer
A. (21)
Concept
Here \(\alpha+\beta=10\) and \(\alpha\beta=q\). Using (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta), (58=100-2q), so (q=21).
Why this answer is correct
The correct answer is A. (21). Here \(\alpha+\beta=10\) and \(\alpha\beta=q\). Using (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta), (58=100-2q), so (q=21).
Exam Tip
यहाँ \(\alpha+\beta=10\) और \(\alpha\beta=q\) है। (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) से (58=100-2q), इसलिए (q=21)।
Login to save your score, XP, coins and progress.