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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 [#permalink]

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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.
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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alexpavlos 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. 7
D. 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.


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:
http://www.purplemath.com/modules/solvquad.htm
http://www.purplemath.com/modules/factquad.htm

Hope it helps.
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 12 May 2012, 08:04
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
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 12 May 2012, 14:14
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!!
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 21 Dec 2012, 06:33
rphardu wrote:
If 4y4 − 41y2 + 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


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
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 26 Dec 2012, 22:58
rphardu wrote:
If 4y4 − 41y2 + 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


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
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 04 Mar 2016, 06:18
\(4y^4 − 41y^2 + 100 = 0\)
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 \(2+\frac{5}{2}\)=\(\frac{9}{4}\)
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]

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New post 04 Mar 2016, 09:14
alex1233 wrote:
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.


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

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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two   [#permalink] 04 Mar 2016, 09:14
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