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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] New post 12 May 2012, 06:23
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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. 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.
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink] New post 12 May 2012, 10:32
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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] New post 12 May 2012, 20:08
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Re: If 4y4 − 41y2 + 100 = 0, then what is the sum of the two gre [#permalink] New post 12 May 2012, 07:04
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] New post 12 May 2012, 13:14
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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If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink] New post 21 Dec 2012, 05:33
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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If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two great [#permalink] New post 26 Dec 2012, 21:58
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] New post 05 Jul 2015, 02:21
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Re: If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two   [#permalink] 05 Jul 2015, 02:21
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