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  • Help me answer: - Complex numbers and quadratic equations - JEE Main-4

Let \alpha ,\beta ,\gamma are three roots of x^{3}+bx^{2}+cx+2b=0 such that \alpha +\beta =0  then ( it is given  b\neq 0 )

  • Option 1)

    c=0

  • Option 2)

    c=1

  • Option 3)

    c=2

  • Option 4)

    c=3

 

Answers (1)

best_answer

\\*\therefore \alpha +\beta +\gamma =-b\; \; but\; \; \alpha +\beta =0\Rightarrow \gamma =-b\\*\because \gamma \;\; is \; \; root\; \; of\; \; equation\; \; so\; \; it\; \; will\; \; satis\! f\! y\; \; the\; \; equation\; \; hence\\*(-b)^{3}+b(-b)^{2}+c(-b)+2b=0\\*\Rightarrow bc=2b\Rightarrow c=2

 

Sum of roots of cubic Equation -

\alpha +\beta +\gamma = \frac{-b}{c}

 

- wherein

ax^{3}+bx^{2}+cx+d= 0

is the cubic equation

 

 


Option 1)

c=0

This is incorrect

Option 2)

c=1

This is incorrect

Option 3)

c=2

This is correct

Option 4)

c=3

This is incorrect

Posted by

Himanshu

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