Question

If beta and 1/beta are zeroes of the polynomial (a^2+a)x^2+61x+6a, find values of Alfa and beta

Answer 1

$\beta + \frac{1}{ \beta } = \frac{ - 61}{ {a}^{2} + a} \\ \beta \times \frac{1}{ \beta } = \frac{6a}{ {a}^{2} + a} \\ \frac{6a}{ {a}^{2} + a} = 1 \\ 6a = {a}^{2} + a \\ {a}^{2} + a - 6a = 0 \\ {a}^{2} - 5a = 0 \\ a(a - 5) = 0 \\ a = 0 \\ a - 5 = 0 \\ a = 5$