The equation x^4+px^3+qx^2+rx+s=0 has two roots which are reciprocols and the other two roots which are opposites.

Show that r=ps.

Question

The equation x^4+px^3+qx^2+rx+s=0 has two roots which are reciprocols and the other two roots which are opposites. 

Show that r=ps.

ganeshie8 · Answer

say the roots are $\alpha, \frac{1}{\alpha }, \beta , -\beta$

ganeshie8 · Answer

$x^4+px^3+qx^2+rx+s=  (x-\alpha )(x-\frac{1}{\alpha })(x-\beta )(x + \beta)$

ganeshie8 · Answer

multiply thru and compare $x^3 , x $ coefficients and constant terms

anonymous · Answer

I did that but it doesn't seem to work out...Let me try get a picture of my working out.

ganeshie8 · Answer

you would get :-
$s = -\beta ^2$

$p = -(\alpha + \frac{1}{\alpha })$

$r  =  \beta^2 ( \alpha + \frac{1}{ \alpha } ) $

ganeshie8 · Answer

so, r = ps.

try again... post the picture il have a look :)

anonymous · Answer

sure, thanks a lot :D by the way, I'm new, how do you give 'medals'?

ganeshie8 · Answer

il message you about that... take a screenshot and npost the work... i cna have a quick look if u want...  :)

anonymous · Answer

attachment

ganeshie8 · Answer

looks good, maybe a small typo somewhere... let me check

ganeshie8 · Answer

got it, for 'x' coefficient, you need to take sum of products of 3 roots at a time

anonymous · Answer

but r=$$\alpha ^{2}\beta$$ not $$\alpha ^{2}\beta+\frac{ \alpha ^{2} }{\beta}$$

anonymous · Answer

doesn't that cancel out to alpha^2*beta

ganeshie8 · Answer

if a, b, c, d are roots, then 

-r = abc + bcd + cda + dab

ganeshie8 · Answer

u need to take product of 3 roots at a time and take their sum, right ?

anonymous · Answer

OH i see what I did wrong I forgot one term hahah!! THANKS A LOT! :)

ganeshie8 · Answer

glad to hear :) nice work btw...  yw !