if a,b and y are zeroes of cubic polynomial x^3+px^2+qx+2 such that ab+1=0.find the value of 2q+q=5
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a, b and y are the zeros of given polynomial .
so,
a^3+pa^2+qa+2=0
b^3+pb^2+qb+2=0
subtracting both side
(a^3-b^3)+(a^2-b^2) p+q (a-b)=0
(a-b)(a^2+b^2+ab)+(a-b)(a+b) p+q (a-b)=0
a=b, and (a^2+b^2+ab)+(a+b) p+q ----(1)
put b = a in equation (1)
(a^2+a^2+a^2)+(a+a) p+q=0
3a^2+2ap+q=0 -------------(2)
given ,
ab+1=0
put here b=a
a^2=-1
now equation (2)
0.3 +2ap+q=0
2ap+q=0
here something is wrong
so,
a^3+pa^2+qa+2=0
b^3+pb^2+qb+2=0
subtracting both side
(a^3-b^3)+(a^2-b^2) p+q (a-b)=0
(a-b)(a^2+b^2+ab)+(a-b)(a+b) p+q (a-b)=0
a=b, and (a^2+b^2+ab)+(a+b) p+q ----(1)
put b = a in equation (1)
(a^2+a^2+a^2)+(a+a) p+q=0
3a^2+2ap+q=0 -------------(2)
given ,
ab+1=0
put here b=a
a^2=-1
now equation (2)
0.3 +2ap+q=0
2ap+q=0
here something is wrong
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