Question

if 2 and -3 are the zeroes of the quadratic polynomial x²+(a+1)x+b,then find the value of a?​

Answer 1

Answer:

$a = 0 \: and \: b = - 6$

Step-by-step explanation:

$let \: f(x) = {x}^{2} + (a + 1)x + b \\ \\ since \: 2 \: and \: - 3 \: are \: the \: zeroes \: of \: f(x) \: so \\ sum \: of \: the \: zeroes = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } \\ 2 + ( - 3) = \frac{ - (a + 1)}{1 } \\ - 1 = - a - 1 \\ - 1 + 1 = - a \\ 0 = - a \\ 0 = a \\ a = 0 \\ \\and \: \\ product \: of \: the \: zeroes = \frac{ constant \: term \: }{coefficient \: of \: {x}^{2} } \\ 2 \times ( - 3) = \frac{b}{1} \\ - 6 = b \\ b = - 6$