3 and 2 are the zeroes of polynomial x^2+px+q.find the value of p and q
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p(x)= x²+px+q
p(3)=0
p(2)=0
so we have
3²+3p+q=0
9+3p+q=0.......……1
2²+2p+q=0
4+2p+q=0…………2
subtracting equation 2 from equation 1
3p+q= -9
-2p-q=+4
= p= -5
putting the value of p in equation 2
-5*-2-q= 4
-10-q= 4
-q= 14
q=14
p(3)=0
p(2)=0
so we have
3²+3p+q=0
9+3p+q=0.......……1
2²+2p+q=0
4+2p+q=0…………2
subtracting equation 2 from equation 1
3p+q= -9
-2p-q=+4
= p= -5
putting the value of p in equation 2
-5*-2-q= 4
-10-q= 4
-q= 14
q=14
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