Question

If the zeroes of the quadratic polynomial x2 + (a+1) x + b are -2 and 3 then find the value of a and b.

Answer 1

Step-by-step explanation:

$given \\ zeroes \: of \: quadratic \: polynmial \: {x}^{2} +( a + 1) x + b \: are \: - 2 \: and \: 3 \\ let \\ p(x) = {x}^{2} + (a + 1)x + b \\ p( - 2) = 0 \\ { (- 2)}^{2} + (a + 1)( - 2) + b = 0 \\ 4 - 2a - 2 + b = 0 \\ 2 - 2a + b = 0 \\ 2a - b =2 = > (1) \\ p(3) = 0 \\ = > {3}^{2} + (a + 1)3 + b = 0 \\ = > 9 + 3a + 3 + b = 0 \\ = > 3a + b = - 12 = > (2) \\ (1) + (2) \\ 2a - b + 3a + b = 2 + ( - 12) \\ 5a = - 10 \\ a = \frac{ - 10}{5} \\ a = - 2 \\ sub \: a = - 2 \: in \: (1) \\ 2( - 2) - b = 2 \\ - 4 - b = 2 \\ b = - 4 - 2 \\ b = - 6 \\ therefore \\ a = - 2 \\ b = - 6$