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

Question

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

A.  4

B.  \frac{9}{2}

C.  7

D.  \frac{41}{4}

E.  25

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.

Bunuel · Accepted Answer

Factor 4y^4-41y^2+100=0 (or just solve for y^2 ) --> (y^2-4)(4y^2-25)=0 : y^2-4=0 --> y=-2 or y=2 ; 4y^2-25=0 --> y=-\frac{5}{2} or y=\frac{5}{2} ; So, the sum of the two greatest possible values of y is 2+\frac{5}{2}=\frac{9}{2} . Answer: B. Solving and Factoring Quadratics: https://www.purplemath.com/modules/solvquad.htm https://www.purplemath.com/modules/factquad.htm Hope it helps.

dabral · Answer

See attached image.

Dabral

ashdah · Answer

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

Gonnaflynow · Answer

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^2-4ac

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!!

mbaiseasy · Answer

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

mbaiseasy · Answer

Let  x = y^2

4(y^2)^2 - 41 (y^2) + 100 = 0 
 4x^2 - 41x^2 + 100 = 0

(4x - 25)(x - 4) = 0 
 x = \frac{25}{4} = y^2 
 y = \frac{5}{2}

x = 4 
 x = y^2 = 4 
 y = 2

Answer:  \frac{5}{2} + 2 = \frac{9}{2}

Answer B

Nevernevergiveup · Answer

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}

chetan2u · Answer

Hi,

every one has generally followed a single method..

I'll just give you two methods incase you get struck..

1) POE-- 
 you can easily eliminate three choices and your prob of answering correctly will go up to 1/2 from 1/5.. 
 4y^4 − 41y^2 + 100 = 0 ..
 41y^2 - 4y^4 =100 ..
 y^2(41 - 4y^2) =100 ..
Now 100 is a positive number, so LHS, y^2(41 - 4y^2), should also be positive..
In y^2(41 - 4y^2), y^2 will always be positive so 41-4y^2>0
or y^2<41/4..
y^2<10.25.. so y will be less than 3.3 approx..
even if both values are 3.3, sum =6.6..
only A and B are left..
 SO , without doing anything, we have eliminated three choices..

2)  4y^4 − 41y^2 + 100 = 0 .. 
lets put this in (a-b)^2 format..
 (2y^2)^2 −2*10*2y^2 + 10^2-y^2 = 0 ..
(2y^2-10)^2=y^2..
so we get two equations..

A) 2y^2-10=y.. 
2y^2-y-10=0..
2y^2-5y+4y-10=0..
y(2y-5) + 2(2y-5)=0..
(y+2)(2y-5)=0..
roots are 5/2 and -2..

B) 2y^2-10=-y.. 
2y^2+y-10=0..
2y^2+5y-4y-10=0..
y(2y+5) - 2(2y-5)=0..
(y-2)(2y+5)=0..
roots are -5/2 and 2..

so values are 5/2, 2, -2, -5/2..
sum of two biggest values= 2+5/2=9/2  
B

AnthonyRitz · Answer

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} .

KanishkM · Answer

I would solve this in this way,

If  4y^4 − 41y^2 + 100 = 0

4y^4 -16y^2 - 25y^2 + 100 = 0

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

y^2 = 25/4 or y^2 = 4

For the roots to be maximum, i will use the +ive roots

5/2 + 2 = 9/2

B

Abhishek009 · Answer

4a^2-41a+100=0  (Where  a = y^2 )

Solve it ,  a = \frac{25}{4}  &  4

Thus,  y = \frac{5}{2}  &  2

So, sum of the two greatest possible values is  \frac{5}{2} + 2 = \frac{9}{2} , Answer will be (B)

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

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

Prep Toolkit

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