85
If $\alpha$ and $\beta$ are the zeroes of the polynomial $ax^2 - x + c$. Obtain a polynomial whose zeroes are $\alpha - 3$ and $\beta - 3$.
Show SolutionHide Solution↓
$\alpha + \beta = \frac{1}{a}, \alpha\beta = \frac{c}{a}$. Sum of zeroes of required polynomial = $\alpha + \beta - 6 = \frac{1}{a} - 6$ or $\frac{1-6a}{a}$. Product of zeroes of required polynomial = $\alpha\beta - 3(\alpha + \beta) + 9 = \frac{c}{a} - \frac{3}{a} + 9$. $\therefore$ required polynomial is $x^2 - (\frac{1-6a}{a})x + \frac{c-3+9a}{a}$ or $ax^2 - (1 - 6a)x + (c - 3 + 9a)$.