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  • Need clarity, kindly explain! - Complex numbers and quadratic equations - JEE Main-8

If sum of roots of equation \left ( a+1 \right )x^{2}+\left ( 2a+3 \right )x+\left ( 3a+4 \right )= 0 is -1, then roots of equation are 

  • Option 1)

    Real & Equal

  • Option 2)

    Real & Distinct

  • Option 3)

    one real & one imaginary

  • Option 4)

    Imaginary

 

Answers (1)

best_answer

\because \: Sum\: =-1\: \Rightarrow \: -\frac{\left ( 2a+3 \right )}{\left (a+1 \right )}=-1\: \Rightarrow \: a=-2

\therefore equation becomes :  -x^{2}-x-2=0

or x^{2}+x+2=0 ;

D=1^{2}-4\left ( 1 \right )\left ( 2 \right )=-7< 0 , so imaginary roots

\therefore Option (D)

 

Sum of Roots in Quadratic Equation -

\alpha +\beta = \frac{-b}{a}

- wherein

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

ax^{2}+bx+c=0

a,b,c\in C

 

 


Option 1)

Real & Equal

This is incorrect

Option 2)

Real & Distinct

This is incorrect

Option 3)

one real & one imaginary

This is incorrect

Option 4)

Imaginary

This is correct

Posted by

prateek

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