niketdoshi123 wrote: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)41/4
d)25
e)7

Source: Kaplan

Use SUBSTITUTION to transform the equation into a typical quadratic.
If z=y², then the equation becomes:
4z² - 41z + 100 = 0.

For any quadratic in the form ax² + bx + c = 0:
The sum of the roots = -b/a.
The product of the roots = c/a.
Thus, the sum of the roots here -(-41)/4 = 41/4.

Since each of the roots is equal to y², and there is no √ in the answer choices, it is virtually guaranteed that each root = (perfect square)/4.
Perfect squares less than 41 are 1,4,9,16,25,36.
Since 16+25 = 41, the two values of y² must be 16/4 and 25/4, yielding a sum of 16/4 + 25/4 = 41/4.

If y² = 16/4 = 4, then y = ±2.
If y² = 25/4, then y = ± 5/2.
Thus, the greatest possible sum = 2 + 5/2 = 9/2.

The correct answer is B.

Not sure there is a better method, but

Taking y^2 as x and solving for the roots gets us 4 and 25/4.
(roots of ax^2 + bx + c = (-b +- rootof(b^2 - 4ac))/ 2a)
These are the possible values for y^2. So, y has to be 2 and
5/2 (taking only +ve roots). Hence sum will be 9/2 => b.

What is the OA ?

Cheers

niketdoshi123 wrote: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)41/4
d)25
e)7

Source: Kaplan