Question

if zeroes of quadratic polynomial X2 +(a+1)x+ b
are 2&-3,then
a> a=-7,b=1


b> a=5, b=-1
c>a=2, b=-6
d>a=0,b =-6​

Answer 1

Answer:

Given that,

${x}^{2} + {a + 1}x + b$

α = 2 & β = -3

Then,

α+β = -b/a

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

- 1 = -(a+1)

a = 0

αβ = c/a

(2)(-3) = b/1

b = -6

Thus, a = 0 and b = -6