Question

find the value of a and b if the zeros of the quadratic polynomial xsquare+(a+1)x+b are 2 and 3 ​

Answer 1

Solution:

Let p(x) = x²+(a+1)x+b

i) 2 is a zero of p(x).

p(2) = 0

=> 2²+(a+1)2+b=0

=> 4+2(a+1)+b =0----(1)

ii) 3 is a zero of p(x).

p(3) = 0

=> 3²+(a+1)3+b = 0

=> 9+(a+1)3+b =0 ----(2)

subtract (1) from (2) , we get

=> 5+a+1=0

=> 6+a = 0

=> a = -6

substitute a=-6 in equation (1),

we get

=> 4+2(-6+1)+b =0

=> 4+2(-5)+b = 0

=> 4-10+b = 0

=> -6+b = 0

=> b =6

Therefore,

a = -6 , b = 6

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