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If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]
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 12 May 2012, 07:23
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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. \(\frac{9}{2}\) 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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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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12 May 2012, 11:32
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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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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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12 May 2012, 21:08
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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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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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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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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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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
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