बहुपद x2 − x − (2 + 2k) का एक शून्यक −4 है, तो k का मान ज्ञात कीजिए। 

This question was previously asked in
Bihar STET Paper I: Mathematics (Held In 2019 - Shift 1)
View all Bihar STET Papers >
  1. 3
  2. 9
  3. 6
  4. −9

Answer (Detailed Solution Below)

Option 2 : 9
Free
Bihar STET Paper 1 Social Science Full Test 1
11.8 K Users
150 Questions 150 Marks 150 Mins

Detailed Solution

Download Solution PDF

प्रयुक्त अवधारणा:

यदि a बहुपद P(x) का शून्यक है तो P(a) = 0

गणना:

−4 बहुपद x2 − x − (2 + 2k) = P(x) का एक शून्यक है। 

अतः, P(-4) = 0

42 - -4 - (2 + 2k) = 0

16 + 4 - (2 + 2k) = 0

20 = (2 + 2k)

20 - 2 = 2k 

18 = 2k, k = 9

अतः, सही उत्तर विकल्प 2 है।

Latest Bihar STET Updates

Last updated on Jul 3, 2025

-> The Bihar STET 2025 Notification will be released soon.

->  The written exam will consist of  Paper-I and Paper-II  of 150 marks each. 

-> The candidates should go through the Bihar STET selection process to have an idea of the selection procedure in detail.

-> For revision and practice for the exam, solve Bihar STET Previous Year Papers.

More Polynomials Questions

More Algebra Questions

Get Free Access Now
Hot Links: teen patti master 2023 teen patti real cash game teen patti neta teen patti app teen patti master 2024