Question

Given that $\alpha$ and $\beta$ are the root of the equation $x^2=7x+4$:
i) Show that ${\alpha }^3=53\alpha +28$
ii) Find the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

Answer 1

$x^2-7x+4$

$⟹x^2-7x-4=0$

$\alpha +\beta =7$

$\alpha \beta =-4$

$1.$$⟹{\alpha }^2-7\alpha -4=0$

$⟹(\alpha -7)({\alpha }^2-7\alpha -4)=0$

$⟹{\alpha }^3-7{\alpha }^2-4\alpha +7{\alpha }^2-49\alpha -28=0$

$⟹{\alpha }^3-53\alpha -28=0$

$⟹{\alpha }^3=53\alpha +28$

$2.$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

$=\frac{-{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{49-2(-4)}{-4}$

$=\frac{49+8}{-4}$

$=-\frac{57}{4}$