यदि n समीकरण x2 + px + m = 0 का एक मूल है और m समीकरण x2 + px + n = 0 का एक मूल है, जहाँ m ≠ n है, तो p + m + n का मान क्या है?

This question was previously asked in
NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)
View all NDA Papers >
  1. -1
  2. 0
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 3 : 1
Free
NDA 01/2025: English Subject Test
5.3 K Users
30 Questions 120 Marks 30 Mins

Detailed Solution

Download Solution PDF

गणना:

दिया गया है,

n समीकरण x2 + px + m = 0 ​का मूल है

\(n^2 + pn + m = 0\) ..... (1)

साथ ही, m समीकरण x2 + px + n = 0 का मूल है

\(m^2 +pm+ n =0\) ..... (2)

अब समीकरण (1) - समीकरण (2), हमें प्राप्त होता है

\((n^2 – m^2) + p(n – m) – (n – m) = 0\)

\((n – m) [(n + m) + p – 1] = 0\)

\(n + m + p – 1 = 0\) ..... ( n ≠ m)

⇒ n + m + p = 1

∴ विकल्प (c) सही है।

Latest NDA Updates

Last updated on May 30, 2025

->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.

-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.

->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.

-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.

-> The selection process for the NDA exam includes a Written Exam and SSB Interview.

-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100. 

-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential. 

More Quadratic Equations Questions

Get Free Access Now
Hot Links: teen patti master 51 bonus teen patti rules teen patti casino teen patti all games teen patti party