PS: Quadratic expression

[amit_agar82]

If 4y4 − 41y2 + 100 = 0, then what is the sum of the two greatest possible values of y ?

4
9/2
7
41/4
25

can someone explain how do I solve 4y4 − 41y2 + 100 = 0

Thanks
Amit

[rajivbhatia2007]

You can try factoring:

(y^2-4)(4y^2-25) = 0

Setting (y^2-4) = 0
y = +/- 2

setting (4y^2-25) = 0
4y^2 = 25
y^2 = 25/4

y = +/- 5/2

so y can be -5/2, -2, 2, 5/2

two largest are 2 and 5/2

the sum is 9/2

i get B.


I realize factoring might be the hardest part.
How I factored this:
Because the coefficient on y^2 is negative and 100 is positive, I realized the factored form would look like
(2y^2 - a)(2y^2 - b)
or
(y^2 - a)(4y^2 - b)

where a and b are positive,
from there i guess and checked.


best,
rajiv

[RonPurewal]

the poster above me has factored this expression correctly. however, if you have extreme trepidation about factoring such expressions, you can always just use the quadratic formula.

here, the quadratic formula yields (41 ± √(1681 - 1600))/8, or (41 ± 9)/8.
that gives 50/8 = 25/4, and 32/8 = 4.
these are the values of y^2, not the values of y, so you have to take ±√ of both of these values to get the possible values of y (which will be the same as those found by the poster above).

--

in any case, i don't know in what source this equation was found, but i have never seen a quadratic this difficult to factor on the actual gmat.
beware of weird third-party sources -- they often demand skill sets far in excess of what is actually required for the official test. in the worst third-party sources, problems are relatively straightforward processes, but with complex or time-consuming operations; that's exactly the opposite of what usually happens on the gmat, on which the problems are very clever and unusual (i.e., not straightforward) but the calculations are, more often than not, fairly elementary.

[amit_agar82]

Thanks Rajiv. That was helpful.

[RonPurewal]

.