If $α$ and $β$ are the zeroes of the polynomial $f (x) = x^{2} - 2 x - 3$ , find the polynomial whose zeroes are $2 α - 1$ and $2 β - 1$ .

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Solution

Since $α, β$ are the zeroes of the polynomial $f (x) = x^{2} - 2 x - 3$
$α + β = - \frac{b}{a} = - (- 2) = 2$
$α β = \frac{c}{a} = - 3$
sum of the zeroes of the required polynomial
$= (2 α - 1) + (2 β - 1) = 2 (α + β) - 2$
$= 2 \times 2 - 2 = 2$
product of the zeroes $= (2 α - 1) (2 β - 1)$
$4 α β - 2 α - 2 β + 1$
$= 4 \times - 3 - 2 (α + β) + 1$
$= - 12 - 2 \times 2 + 1 = - 15$
sum of zeroes $= 2 = - \frac{b}{a}$
product of zeroes $= - 15 = \frac{c}{a}$
If $a = 1$ , then $b = - 2, c = - 15$ in $a x^{2} + b x + c$
The required polynomial is $x^{2} - 2 x - 15$

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If α and β are the zeroes of the polynomial f(x)=x2−2x−3, find the polynomial whose zeroes are 2α−1 and 2β−1.

If $α$ and $β$ are the zeroes of the polynomial $f (x) = x^{2} - 2 x - 3$ , find the polynomial whose zeroes are $2 α - 1$ and $2 β - 1$ .