If the zeroes of the quadratic polynomial x2+(a+1)x+b are 2 and -3, then find the value of a and b.
Answers
Answered by
4
If one zero = 2
p(x) = x2 +(a+1) x + b
So, if x=2
p(2)=0
2^2 + ( a+1)2+b=0
=> 4+2a+ 2 + b =0
=> 2a+ b = -2. ————-Eq 1
If x= -3
=> (-3)^2 +(a+1)(-3) + b =0
=> 9 -3a-3 +b =0
=> -3a+b= -6 —————Eq2
On solving both equations u get a= 4/5 and b = -18/5
p(x) = x2 +(a+1) x + b
So, if x=2
p(2)=0
2^2 + ( a+1)2+b=0
=> 4+2a+ 2 + b =0
=> 2a+ b = -2. ————-Eq 1
If x= -3
=> (-3)^2 +(a+1)(-3) + b =0
=> 9 -3a-3 +b =0
=> -3a+b= -6 —————Eq2
On solving both equations u get a= 4/5 and b = -18/5
Similar questions