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  • Please,please help me - Complex numbers and quadratic equations - JEE Main-5

Let \alpha ,\beta are two roots of x^{3}+px^{2}+qx+r=0 & satisfies \alpha \beta =-1,  if r \neq 0  then \left | r^{2}+pr+q \right |  equals

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Answers (1)

\\*\alpha \beta \gamma =-r\\*\because \alpha \beta =-1\Rightarrow \gamma =r\\*\therefore\; it\; \; will\; \; satisfy\; \; \; the\; \; equation\\*\Rightarrow r^{3}+pr^{2}+qr+r=0\\*\Rightarrow r^{2}+pr+q=-1\\*\therefore \left | r^{2}+pr+q \right |=1

 

Product of roots of cubic equation -

\alpha \beta \gamma = \frac{-d}{a}

- wherein

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

is the cubic equation

 

 


Option 1)

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Option 4)

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