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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

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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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