Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Give answer! - Complex numbers and quadratic equations - JEE Main-6

If one root of 5x^{2}+13x+k= 0 is reciprocal of other, then equation has 

  • Option 1)

    Real & Distinct roots

  • Option 2)

    Real & Equal roots 

  • Option 3)

    Imaginary roots

  • Option 4)

    one real & one imaginary root

 

Answers (1)

best_answer

\because roots are reciprocal, So their product = 1

\therefore \: \frac{k}{5}=1\: \Rightarrow \: k=5

\therefore equation becomes :-

5x^{2}+13x+5=0

Which has D=13^{2}-4\left ( 5 \right )\left ( 5 \right )

                        =69> 0

\therefore roots are real and distinct

\therefore Option (A)

 

Product of Roots in Quadratic Equation -

\alpha \beta = \frac{c}{a}

- wherein

\alpha \: and\ \beta are roots of quadratic equation:

ax^{2}+bx+c=0

a,b,c\in C

 

 

 


Option 1)

Real & Distinct roots

This is correct

Option 2)

Real & Equal roots 

This is incorrect

Option 3)

Imaginary roots

This is incorrect

Option 4)

one real & one imaginary root

This is incorrect

Posted by

Plabita

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

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

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question