यदि (x=0) समीकरण \(2x^2+mx+n=0\) का मूल है तो (n) का मान क्या होगा?
If (x=0) is a root of \(2x^2+mx+n=0\), what will be the value of (n)?
Explanation opens after your attempt
A. (0)
Concept
Putting (x=0) leaves only (n). Therefore (n=0) must hold.
Why this answer is correct
The correct answer is A. (0). Putting (x=0) leaves only (n). Therefore (n=0) must hold.
Exam Tip
(x=0) रखने पर केवल (n) बचता है। इसलिए (n=0) होना चाहिए।
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