 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Same Real Roots (Posted on 2016-12-10) There exists a polynomial of the form

x^6 + ax + b

that has the same set of real roots as the polynomial

x^2 - 2x - 1

Find |a+b|

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 2
if x^6+ax+b has the same roots of x^2-2x-1 then x^2-2x-1 is a factor of x^6+ax+b
thus
x^6+ax+b=(x^2-2x-1)(x^4+px^3+qx^2+rx+s)
for some values of p,q,r,s
expanding and collecting on terms of x we get
x^6+ax+b=x^6+(p-2)x^5+(q-2p-1)x^4+(r-2q-p)x^3+(s-2r-q)x^2+(-2s-r)x-s
from the x^5 terms we have
p-2=0 giving p=2
from the x^4 terms we have
q-2p-1=0
q-5=0
q=5
from the x^3 terms we have
r-2q-p=0
r-12=0
r=12
from the x^2 terms we have
s-2r-q=0
s-29=0
s=29

this gives us
a=-2s-r=-70
b=-s=-29

thus we have
x^6+ax+b=x^6-70x-29=(x^2-2x-1)(x^4+2x^3+5x^2+12x+29)
and the quartic factor has no real roots
thus x^6-70x-29 has the same real roots as x^2-2x-1

thus
|a+b|=|-70-29|=|-99|=99

 Posted by Daniel on 2016-12-10 09:04:05 Please log in:
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