Question

In quadratic equation 2y = 10 - y^2. FFind the value of b?​

Answer 1

Step-by-step explanation:

$2y = 10 - {y}^{2} \\ = > {y}^{2} + 2y - 10 = 0 \\ here \: a = 1 \: b = 2 \: and \: c = - 10 \\ then \\ discriminant(d) = \sqrt{ {b}^{2} - 4ac } \\ = \sqrt{ {2}^{2} - 4 \times 1 \times ( - 10) } \\ = \sqrt{4 + 40} = \sqrt{44} \\ then \: roots \: of \: y \: are \\1st > \frac{ - b + \sqrt{d} }{2a} = \frac{ - 2 + \sqrt{44} }{2} \\ = - 1 + \sqrt{11} \\ 2nd > \frac{ - b - \sqrt{d} }{2a} = \frac{ - 2 - \sqrt{44} }{2} \\ = - 1 - \sqrt{11}$