Question

If alfa and beeta are the zeroes of the polynomial 6y - 7y + 2, find a quadratic poly whose zeroes are 1 upon alfa and 1 upon beeta

Answer 1

$6 {y}^{2} - 7y + 2 \\ \alpha + \beta = \frac{7}{6} \\ \alpha \beta = \frac{2}{6} = \frac{1}{3}$
$sum \: of \: zeros \: of \: new \\ \: polynomial \: \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } \\ = \frac{ \frac{7}{6} }{ \frac{1}{3} } \\ = \frac{7 \times 3}{6} = \frac{7}{2} = \frac{ - b}{a} \\ product \: of \: zeros \\ \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{ \alpha \beta } = \frac{1}{ \frac{1}{3} } = 3 \\ so \: new \: polynomials \: will \: be \\ {y}^{2} - \frac{7}{2} y + 3 = 0 \\ 2 {y}^{2} - 7y + 6 = 0 \: \: \: \: ans$