Question

The zeroes of x2+(a+1)x+b are 2 and -3 find a and b

Answer 1

Step-by-step explanation:

Sum of zeroes = -b/a

2 + (-3) = -(a + 1)

2 - 3 = -a - 1

-1 = -a - 1

a = -1 + 1

a = 0

Now , product of zeroes = c/a

2(-3) = b/1

-6 = b

b = -6

Answer 2

Given equation

$x^{2} +(a+1)x+b=0$

2 and -3 are the zeroes of the polynomial

Substituting 2 in the equation, we get,

$\implies (2)^{2} +(a+1)2+b=0$

$\implies 4+2a+2+b=0$

$\implies 2a+b+6=0$

$\implies 2a+b=-6--(1)$

Substituting -3 in the equation, we get,

$\implies (-3)^{2} +(a+1)(-3)+b=0$

$\implies 9-3a-3+b=0$

$\implies -3a+b+6=0$

$\implies -3a+b=-6--(2)$

Solving (1) and (2) we get,

$2a+b=-6\\\underline{-3a+b=-6}\\\underline{\underline{5a=0}}\\\implies a = 0\\\implies b = -6$

$\boxed{\boxed{\bold{Therefore, \ the \ values \ of \ a \ and \ b \ are \ 0 \ and \ -6 \ respectively}}}}}}$