if one zero of polynomial is (a square + 9) X square + 13 x + 6a is reciprocal of the Other find the value of a
Answers
Answered by
5
Let the roots of (a^2+9)x^2+13x+6a be p and 1/p
Product of roots = p×1/p=1 =constant term/coefficient of x^2
6a/(a^2+9)=1
6a=a^2+9
a^2+9-6a=0
a^2-6a+9=0
a^2-3a-3a+9=0
a(a-3)-3(a-3)=0
(a-3)(a-3)=0
(a-3)^2=0
a-3=0
a=3
Product of roots = p×1/p=1 =constant term/coefficient of x^2
6a/(a^2+9)=1
6a=a^2+9
a^2+9-6a=0
a^2-6a+9=0
a^2-3a-3a+9=0
a(a-3)-3(a-3)=0
(a-3)(a-3)=0
(a-3)^2=0
a-3=0
a=3
Similar questions