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How to solve this problem- - Complex numbers and quadratic equations - JEE Main

If  \alpha ,\beta  are roots of  x^{2}+px+q= 0 and  \gamma ,\delta  are roots of x^{2}+px-r=0 then \left ( \gamma -\alpha \right )\left ( \gamma -\beta \right )  equals

  • Option 1)

    q+r

  • Option 2)

    q-r

  • Option 3)

    r-q

  • Option 4)

    0

 
Answers (1)
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x^{2}+px+q=\left ( x-a \right )\left ( x-b \right )

\Rightarrow \: \left ( \gamma -\alpha \right )\left ( \gamma -\beta \right )=\gamma ^{2}+p\gamma +q\cdots \cdots \left ( 1 \right )

\therefore \: \gamma is root of  x^{2}+px-r\cdots \cdots \left ( 2 \right )

From (1) and (2) we get

\left ( \gamma -\alpha \right )\left ( \gamma -\beta \right )=q+r

\therefore Option (A)

 

Factor Theorem -

Any polynomial can be written in terms of product of its factors.

- wherein

If P\left ( x \right )= 0  has roots  \alpha _{1},\alpha _{2},\cdots \alpha _{n}  and  P\left ( x \right )  has leading coefficient 'a' then then P\left ( x \right )= a\left ( x-\alpha _{1} \right )\left ( x-\alpha _{2} \right )\cdots \left ( x-\alpha _{n} \right )  

 

 


Option 1)

q+r

This is correct

Option 2)

q-r

This is incorrect

Option 3)

r-q

This is incorrect

Option 4)

0

This is incorrect

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