Question

If one zero of the quadratic polynomial [(K^2 + 9)x^2 + 13x + 6k) is reciprocal of
the other, find k.
​

Answer 1

Answer:

k = 3

Step-by-step explanation:

product of zeros

$\alpha \beta = x \times \frac{1}{x} = 1 \\ \alpha \beta = \frac{c}{a} \\ = \frac{6k}{ {k}^{2} + 9} = 1 \\ 6k = {k}^{2} + 9 \\ {k}^{2} - 6k + 9 = 0 \\ {k}^{2} - 3k - 3k + 9 \\ k(k - 3) - 3(k - 3) \\ k - 3 = 0 \: \: \: k - 3 = 0 \\ k = 3$