Question

If 2 and -3 are the zeroes of polynomial x^2-(a+1)x+b then find the value of a and b

Answer 1

Step-by-step explanation:

Solution:-

Given 2 and -3 are the zeroes of polynomial

f(×)=x²-(a+1)x+b

first take x=2 , we get

now put the value of x=2 on given polynomial :-x²-(a+1)x+b

f(2)=(2)²-(a+1)×2+ b

0=4-2a+2+b

0=6-2a+b

2a-b=6 ..............(i)eq

now take x= -3

put the value of x= -3 on given polynomial:-x²-(a+1)x+b

f(-3)= (-3)²-(a+1)×-3+b

0=9-a×(-3)+1×-3 +b

0=9+3a-3+b

0= 6+3a+b

3a+b= -6 ................(ii) eq

we have two unknown value and two equation

using substitution method

2a-b=6 .........(i) eq

3a+b= -6 .........(ii) eq

now find the value of b from i st equation

2a-b=6

-b=6-2a

b= -(6-2a)

put the value of b on (ii) eq

3a+b = -6

3a-6+2a= -6

5a= -6+6

5a =0

a =0

now put the value of a=0 on b,we ger

b= -6+2a

b= -6+2×0

b= -6

Answer:- a=0 and b= -6