Question

the roots of the polynomial equation 2x^3-8x^2+3x+5=0 are a,b,c. Find the polynomial equation with roots a^2,b^2,c^2.
Also, how do you answer these type of questions where they ask you to find different versions of the roots?

Answer 1

http://openstudy.com/users/hartnn#/updates/505c45fee4b02e13940f9551

similar to this

what have you tried yet @duplicitycheese ?

Answer 2

here let y=a^2 hence a=sqrt(y)
since a is the root of the given equation so a satisfies it hence we r to replace x by sqrt(y)
in the above equations and square the terms if(necessary ) to make y free from surds

Answer 3

\[\large{a+b+c=0}\]

Answer 4

Hm, well I've tried fiddling with the product of the roots - squaring it and rearranging it... but it looked too weird

Answer 5

@hartnn

Answer 6

' hence we r to replace x by sqrt(y)'
its correct,show your work

Answer 7

but maybe not the best approach here...
i can see that becoming complicated

Answer 8

Oh, I shall try that.

Answer 9

good spirit @duplicitycheese

Answer 10

hmm...that approach is not difficult.....if u don't get it, ask...

Answer 11

I got stuck after subbing in sqrt(y). Do I square the whole thing?

Answer 12

i will get u to that step.....
\(\large 2y^{3/2}-8y+3y^{1/2}+5=0 \\ \large8y-5=y^{1/2}(2y+3)\)
Now square both sides.......

Answer 13

squaring...

Answer 14

ok, got it but there's still sqrts in here:
4y^3-32y^(5/2)+64y^2=9y+30y^(1/2)+25

Answer 15

u did it incorrectly....
(8y-5)^2 = y (2y+3)^2
where does y^(1/2) come from ??

Answer 16

ah opps, sorry I squared the wrong things.

Answer 17

Is this the right answer? 4y^3-32^y2+89y-25?

Answer 18

*52y^2

Answer 19

4y^3 -52y^2+89y-25=0

Answer 20

yay! Thanks!

Answer 21

welcome :)