Question

if zeroes of quadratic equation is 2, -3 then find a and b of the given equation x^2+(a+1)x+b​

Answer 1

Answer:

Step-by-step explanation:

given 2,-3 are the zeros of the equation

now

p(2)=2^2+(a+1)2+b=0

4+2a+2+b=0

2a+b+6=0---(1)

p(-3)=(-3^2)+(a+1)(-3)+b=0

9-3a-3+b=0

-3a+b=-6

3a-b=6---(2)

(1) and (2) solve

5a=0

a=0

b=-6