Question

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

Answer 1

sum of zeroes = -(a+1) = 2+(-3) =-1
a+1 = 1
a= 0

product of zeoes = b = (2)(-3)
b = -6

Answer 2

I was quite perplexed too so here goes:

Answer:

a=0; b=-6

Step-by-step explanation:

x² + (a+1)x + b

2 and -3 are zeroes of the polynomial,

put x = 2

» 2² + (a+1)(2) + b = 0

» 4 + 2a+2 + b = 0

» 6 + 2a+b = 0

» 2a+b = -6 -----(1)

put x = -3,

» (-3)² + (a+1)(-3) + b = 0

» 9 - 3a-3 + b = 0

» 6 - 3a+b = 0

» -3a+b = -6 -----(2)

(1) - (2)

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

2a+b+3a-b = -6+6

5a = 0

a = 0

________

2a + b = -6

2(0)+b = -6

b = -6