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

4y^4 − 41y^2 + 100 = 0

Let y^2=x

4x^2-41x+100=0

by finding the roots of the equation we get using [(-b(+/_)[square_root]b^2-4ac[/square_root)/2a}]

x=4 or 25/4

so y=2 or -2 or y=5/2 or -5/2

so adding the positive values

=2+5/2=9/2

Hence B

Hope that helps!!
_________________

Work with hope in Heart and dreams in the eyes .... And leave the mind for GMAT problems

Let y^2 = x

4x^2 - 41x + 100 = 0
(4x - 25)(x - 4) = 0
x = 25/4 or x = 4

y = 5/2 or y = 2

= 5/2 + 2 = 9/2

Answer: B
_________________

Impossible is nothing to God.

\(4y^4 − 41y^2 + 100 = 0\)
This can be represented in the form of \((a^2-2ab+b^2)=(a-b)^2\) by adding and subtracting \(b^2\) here for convenience.

\((2y^2)^2-2(2y^2)\frac{41}{4}+(\frac{41}{4})^2-(\frac{41}{4})^2+100=0\)

\((2y^2-\frac{41}{4})^2+100-(\frac{41}{4})^2=0\)

\((2y^2-\frac{41}{4})^2-(\frac{81}{16})=0\)

\((2y^2-\frac{41}{4})=\sqrt{(\frac{81}{16})}\)

\((2y^2-\frac{41}{4})\)=+\((\frac{9}{4})\)

\(2y^2=\frac{41}{4})\)+\((\frac{9}{4})\)

\(2y^2=\frac{50}{4} or \frac{32}{4}\)

Thus \(y=\)+\(\frac{5}{2}\) or +\(2\)

\(y=\frac{5}{2},\frac{-5}{2}\),\(+2\), \(-2\)

The two greatest possible values of \(y\) are \(+2\) and \(\frac{+5}{2}\)

their sum is \(2+\frac{5}{2}\)=\(\frac{9}{4}\)

Cheatyface method (for people too lazy to factor):

Divide through by \(4\) to obtain \(y^4 − \frac{41}{4}y^2 + 25 = 0\).

Substitute \(A = y^2\) to obtain \(A^2 − \frac{41}{4}A + 25 = 0\).

We know that the two solutions to a quadratic of the form \(x^2+bx+c = 0\) must add up to \(-b\), so our two solutions for \(A\), a.k.a. \(y^2\), add to \(\frac{41}{4}\).

Of course the \(\frac{41}{4}\) answer choice is for suckers; we want the sum of the positive roots of these two solutions. And we note that the answer choices are all rational numbers. So there ought to be two perfect squares that add to \(\frac{41}{4}\). How about \(\frac{25}{4}\) and \(\frac{16}{4}\)? (Bonus: These values correctly multiply out to \(25\).) Well okay, the positive roots here are \(\frac{5}{2}\) and \(\frac{4}{2}\), and they add to \(\frac{9}{2}\).