Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, - Complex numbers and quadratic equations - JEE Main

The value of 'a' for which x^{2}+2\left ( a-1 \right )x+\left (a+5 \right )= 0 have roots of opposite sign is

  • Option 1)

    \left ( -\infty,-5 \right )

  • Option 2)

    (-\infty,-5]

  • Option 3)

    \left ( -5,\infty \right )

  • Option 4)

    [-5,\infty)

 

Answers (1)

D> 0 and \frac{c}{a}< 0

 

\Rightarrow \: 4\left ( a^{2}-2a+1 \right )-4\left ( a+5 \right )> 0  &  \frac{a+5}{1}< 0

\Rightarrow \: a^{2}-3a-4> 0 & a< -5

\Rightarrow \: \left ( a-4 \right )\left ( a+1 \right )> 0 & a< -5

\therefore \: a< -1\, \cup \, a> 4 & a< -5

\therefore a< -5

\therefore  Option (A)

 

Roots are of opposite sign -

\frac{c}{a}< 0,D= b^{2}-4ac> 0

-

 

 


Option 1)

\left ( -\infty,-5 \right )

This is correct

Option 2)

(-\infty,-5]

This is incorrect

Option 3)

\left ( -5,\infty \right )

This is incorrect

Option 4)

[-5,\infty)

This is incorrect

Posted by

Vakul

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

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question