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If $\alpha$ and $\beta$ are the roots of the equation $(x-a)(x-b)=c, c \neq 0$; then the roots of the equation $(x-\alpha)(x-\beta)+c=0$ are
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Solution:
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Verified Answer
The correct answer is:
$a, b$
Option 3 a,b
Concept:
Quadratic equations whose roots are $\alpha$ and $\beta$ is given by
$(x-\alpha)(x-\beta)$
Calculation:
Given that,
$(x-a)(x-b)-c$ ...(i)
According to the question, $\alpha$ and $\beta$ are the roots of the equation.
Therefore,
$\begin{aligned} & (x-a)(x-b)-c=(x-\alpha)(x-\beta) \\ & \Rightarrow(x-a)(x-b)=(x-\alpha)(x-\beta)+c\end{aligned}$
which represents, root of equation $(x-\alpha)(x-\beta)+c$ are $a$ and $b$.
Concept:
Quadratic equations whose roots are $\alpha$ and $\beta$ is given by
$(x-\alpha)(x-\beta)$
Calculation:
Given that,
$(x-a)(x-b)-c$ ...(i)
According to the question, $\alpha$ and $\beta$ are the roots of the equation.
Therefore,
$\begin{aligned} & (x-a)(x-b)-c=(x-\alpha)(x-\beta) \\ & \Rightarrow(x-a)(x-b)=(x-\alpha)(x-\beta)+c\end{aligned}$
which represents, root of equation $(x-\alpha)(x-\beta)+c$ are $a$ and $b$.
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