यदि \(\alpha,\beta\) समीकरण \(7x^2-25x+12=0\) के मूल हैं, तो \(\alpha+\beta\) क्या है?
If \(\alpha,\beta\) are roots of \(7x^2-25x+12=0\), what is \(\alpha+\beta\)?
Explanation opens after your attempt
Correct Answer
A. \( \frac{25}{7}\)
Step 1
Concept
The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{25}{7}\). The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें।
Mathematics Answer, Explanation and Revision Hints
यदि \(\alpha,\beta\) समीकरण \(7x^2-25x+12=0\) के मूल हैं, तो \(\alpha+\beta\) क्या है? / If \(\alpha,\beta\) are roots of \(7x^2-25x+12=0\), what is \(\alpha+\beta\)?
Correct Answer: A. \( \frac{25}{7}\). Explanation: मूलों का योग \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें। / The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
Which concept should I revise for this Mathematics MCQ?
The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
What exam hint can help solve this Mathematics question?
मूलों का योग \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें।
Login to view answers
Use Google login or mobile OTP. Admin can block users and control login providers.
Student Class Required
Select your class first
Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.
