Question

If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and –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:

a=0,b=-6

Step-by-step explanation:

x=2

Hence, x²+(a+1)x+b=0

=>(2)²+(a+1)2+b=0

=> 4+2a+2+b=0

=> 6+2a+b =0

=> 2a+b =-6 ------------equation (1)

x=-3

Hence,x²+(a+1)x+b=0

=>(-3)²+[-3(a+1)]+b=0

=> 9+[-3a+(-3)]+b=0

=> 9-3a-3+b=0

=> 6-3a+b=0

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

Now ,subtract equation (2) from (1).

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

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

5a=0

a=0

Then,from equation (1) we get:-

2a+b=-6

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

Now substituting the value of a in equation (3) we get:-

b=-6-2×0

b=-6

Hence ,a=0 and b=-6