Question
If a is a root for the following equation
x^2+\sqrt{3}x-2 = 0
Find the value of
\dfrac{a^4}{a^8+a^6+2a^4+4a^2+16}
Answer
Since a is one of roots of the equation x^2+\sqrt{3}x-2 = 0,
a^2+\sqrt{3}a-2 = 0
a^2-2 = -\sqrt{3}a
a-\dfrac{2}{a} = \sqrt{3}
(1)
Taking square of the equation
a^2-4+\dfrac{4}{a^2} =3
a^2+\dfrac{4}{a^2} = 7
(2)
Taking square of the equation (2) gives
a^4+\dfrac{16}{a^4} = 41
(3)
Divide both numerator and denominator by a^4, and substitute results of (2) and (3)
\dfrac{a^4}{a^8+a^6+2a^4+4a^2+16}
=\dfrac{1}{a^4+a^2+2+\dfrac{4}{a^2}+\dfrac{16}{a^4} }
=\dfrac{1}{a^4+\dfrac{16}{a^4}+9 }
=\dfrac{1}{50}