Other Than Golden Ratio?

Algebra Level 3

Suppose $\alpha$ and $\beta$ are the two roots to $x^2-x-1$ , find the value of $\alpha^4+3\beta$ .

1 solution

Brian Charlesworth
Feb 13, 2015

Given that $\alpha$ is a solution to the equation $x^{2} - x - 1 = 0$ , we know that

$\alpha^{2} = \alpha + 1 \Longrightarrow \alpha^{4} = \alpha^{2} + 2\alpha + 1 = 3\alpha + 2.$

So $\alpha^{4} + 3\beta = 2 + 3(\alpha + \beta).$ But by Vieta's rule we know that $\alpha + \beta = 1$ , and so

$2 + 3(\alpha + \beta) = \boxed{5}.$

Sir you made it really simple. I used a very long way.

$x^2 - x - 1 =0$

$\implies x = \dfrac{1 \pm \sqrt{5}}{2}$

$\implies x = \varphi , \psi$

$\varphi^2 - 1 = \varphi$

$\alpha^4 + 3\beta = \alpha^4 - 1 + 3\beta + 1$

$= \alpha(\alpha^2 + 1) + 3\beta + 1$

$= \sqrt{5}( \alpha)^2 + 3\beta + 1$

$= \sqrt{5}( \alpha^2 - 1) + 3\beta + 1 + \sqrt{5}$

$= 5$

U Z - 6 years, 3 months ago

yep, i did the same way. Brian sir is Awesome.

Keshav Tiwari - 6 years, 3 months ago