Hard Mathematics Quadratic Equations Class 10 Level 31

यदि (x-2-2(m+1)x+\(m^2+2m+1\)=0) के मूल बराबर हैं तो (m) के लिए क्या सही है?

If roots of (x-2-2(m+1)x+\(m^2+2m+1\)=0) are equal, what is true for (m)?

Explanation opens after your attempt
Correct Answer

A. हर वास्तविक (m)Every real (m)

Step 1

Concept

The constant term is ((m+1)2), and the equation becomes ((x-(m+1))2=0). Hence roots are equal for every real (m).

Step 2

Why this answer is correct

The correct answer is A. हर वास्तविक (m) / Every real (m). The constant term is ((m+1)2), and the equation becomes ((x-(m+1))2=0). Hence roots are equal for every real (m).

Step 3

Exam Tip

अचर पद ((m+1)2) है और समीकरण ((x-(m+1))2=0) बनता है। इसलिए हर वास्तविक (m) के लिए मूल बराबर हैं।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (x-2-2(m+1)x+\(m^2+2m+1\)=0) के मूल बराबर हैं तो (m) के लिए क्या सही है? / If roots of (x-2-2(m+1)x+\(m^2+2m+1\)=0) are equal, what is true for (m)?

Correct Answer: A. हर वास्तविक (m) / Every real (m). Explanation: अचर पद ((m+1)2) है और समीकरण ((x-(m+1))2=0) बनता है। इसलिए हर वास्तविक (m) के लिए मूल बराबर हैं। / The constant term is ((m+1)2), and the equation becomes ((x-(m+1))2=0). Hence roots are equal for every real (m).

Which concept should I revise for this Mathematics MCQ?

The constant term is ((m+1)2), and the equation becomes ((x-(m+1))2=0). Hence roots are equal for every real (m).

What exam hint can help solve this Mathematics question?

अचर पद ((m+1)2) है और समीकरण ((x-(m+1))2=0) बनता है। इसलिए हर वास्तविक (m) के लिए मूल बराबर हैं।

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.