If $\alpha, \beta$ are the roots of $2x^2+5x+1=0$, then the equation whose roots are $2\alpha +1, 2\beta +1$ is

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Solution

$ f(x) = 2x^2+5x+1 = 0 \rightarrow \alpha, \beta $ are the roots.
Let $y = 2x+1$, where $x =\alpha,\beta$
$x = \dfrac{y-1}{2}$
The equation whose roots are $ 2 \alpha +1, 2\beta +1$ is,
$f \left(\dfrac{x-1}{2}\right)= 0 $
$\Rightarrow \dfrac{2 (x-1)^2}{4}+ \dfrac{5(x-1)}{2}+1 = 0$
$\Rightarrow (x-1)^2 +5 (x-1)+2 = 0\\
\Rightarrow x^2+ 3x-2 = 0$

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