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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)

best_answer

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

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

$\therefore \: \gamma$ 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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