alexpavlos wrote:

I get stuck when trying to solve this one. I can see that it will end up something along the lines of (x^2)^2=x^2 to factor it but still struggling.

If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two greatest possible values of y ?

A) 4

B) 9/2

C) 7

D) 41/4

E) 25

Hi this can be solved as below:-

we have

4y^4 − 41y^2 + 100 = 0

that can be factorised as (a-b)^2=a^2+b^2-2ab------------------------(a)

so we have (2y^2)^2-(2*(2y^2)(10))+10^2-y^2=0

or using (a)

(2y^2-10)^2-2y^2=0

or (2y^2-10)^2=2y^2

i.e we have two solns

on taking square root on both sides

(2y^2-10)=2y-----------------------(b)

or (2y^2-10)=-2y-----------------(c)

on solving (b) as normal eqn we have

(2y-5)(y+2) =0

so max value is y =5/2

on solving (c) we have

(2y+5)(y-2) =0

or max value as y=2

so adding these two values we have

2+5/2==9/2

I hope this helps