Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me understand! - Complex numbers and quadratic equations - JEE Main-4

Let \alpha ,\beta ,\gamma are roots of x^{3}+x+1= 0 then the equation with roots \alpha -1,\beta -1,\gamma -1 is 

  • Option 1)

    x^{3}+3x^{2}+4x+3= 0

  • Option 2)

    x^{3}-3x^{2}+4x+3= 0

  • Option 3)

    x^{3}-3x^{2}-4x+3= 0

  • Option 4)

    x^{3}-3x^{2}+4x-3= 0

 

Answers (1)

best_answer

Let \alpha -1=y\: \Rightarrow \: \alpha =y+1 , Now putting \alpha in given equation we get - 

\left ( y+1 \right )^{3}+\left ( y+1 \right )+1=0\: \Rightarrow \: y^{3}+3y^{2}+4y+3=0

\therefore Equation is \rightarrow \: x^{3}+3x^{2}+4x+3=0

\therefore  Option (A)

 

Transformation of equation -

To find equation whose roots are symmetrical functions of \alpha and \beta , Where \alpha & \beta are roots of some other equation. 

- wherein

Take any of the root to be equal to y  &  calculate \alpha  or  \beta  accordingly in terms of y  & satisfy the given equation to get required equation.

 

 


Option 1)

x^{3}+3x^{2}+4x+3= 0

This is correct

Option 2)

x^{3}-3x^{2}+4x+3= 0

This is incorrect

Option 3)

x^{3}-3x^{2}-4x+3= 0

This is incorrect

Option 4)

x^{3}-3x^{2}+4x-3= 0

This is incorrect

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question