Muft ShikshaQuiz Club
If (x=-4) is a root of \(2x^2+px-12=0\), what is the value of (p)?
A. (5)
Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
The correct answer is A. (5). Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
(x=-4) रखने पर (32-4p-12=0) मिलता है। इससे (p=5) है।
Login to save your score, XP, coins and progress.
Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365
Login karke aap is question ka doubt, correction ya suggestion bhej sakte hain aur reply track kar sakte hain.
Correct Answer: A. (5). Explanation: (x=-4) रखने पर (32-4p-12=0) मिलता है। इससे (p=5) है। / Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
(x=-4) रखने पर (32-4p-12=0) मिलता है। इससे (p=5) है।
Use Google login or mobile OTP. Admin can block users and control login providers.
Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.