If $a_{1}$ and $a_{2}$ are the zeros of the quadratic polynomial $p (y) = y^{2} - y - 2$ then find the polynomial whose zeros are $2 a_{1} + 1$ and and $2 a_{2} + 1$ .

k [y^{2} - 4 y - 5]

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k [y^{2} + 4 y - 5]

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k [y^{2} + 4 y + 5]

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k [y^{2} - 4 y + 5]

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Solution

The correct option is A $k [y^{2} - 4 y - 5]$

Since $a_{1}$ and $a_{2}$ are zeroes of
$p (y) = y^{2} - y - 2$
$a_{1} + a_{2} = 1$ and $a_{1} \times a_{2} = - 2$
Now polynomial with zeroes $2 a_{1} + 1$ and $2 a_{2} + 1$ is given by
$y^{2} - [(2 a_{1} + 1) + (2 a_{2} + 1)] y + (2 a_{1} + 1) (2 a_{2} + 1) = y^{2} - [2 (a_{1} + a_{2}) + 2] y + 4 a_{1} a_{2} + 2 a_{1} + 2 a_{2} + 1 = y^{2} - [2 \times 1 + 2] y + 4 \times (- 2) + 2 (1) + 1 = y^{2} - 4 y - 8 + 3 = y^{2} - 4 y - 5$
Thus, the required polynomial is $k [y^{2} - 4 y - 5]$ .

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