#### 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}

Collected in the board: Nike function

Steven Zheng posted 4 minutes ago

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}

Steven Zheng posted 4 minutes ago

Scroll to Top