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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 (ab)^2=a^2+b^22ab(a) so we have (2y^2)^2(2*(2y^2)(10))+10^2y^2=0 or using (a) (2y^210)^22y^2=0 or (2y^210)^2=2y^2 i.e we have two solns on taking square root on both sides (2y^210)=2y(b) or (2y^210)=2y(c) on solving (b) as normal eqn we have (2y5)(y+2) =0 so max value is y =5/2 on solving (c) we have (2y+5)(y2) =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
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4y^4 − 41y^2 + 100 = 0 Let y^2=x 4x^241x+100=0 by finding the roots of the equation we get using [(b(+/_)[square_root]b^24ac[/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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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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04 Mar 2016, 06:18
\(4y^4 − 41y^2 + 100 = 0\) This can be represented in the form of \((a^22ab+b^2)=(ab)^2\) by adding and subtracting \(b^2\) here for convenience. \((2y^2)^22(2y^2)\frac{41}{4}+(\frac{41}{4})^2(\frac{41}{4})^2+100=0\) \((2y^2\frac{41}{4})^2+100(\frac{41}{4})^2=0\) \((2y^2\frac{41}{4})^2(\frac{81}{16})=0\) \((2y^2\frac{41}{4})=\sqrt{(\frac{81}{16})}\) \((2y^2\frac{41}{4})\)= +\((\frac{9}{4})\) \(2y^2=\frac{41}{4})\) +\((\frac{9}{4})\) \(2y^2=\frac{50}{4} or \frac{32}{4}\) 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 414y^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 (ab)^2 format.. \((2y^2)^2 −2*10*2y^2 + 10^2y^2 = 0\).. (2y^210)^2=y^2.. so we get two equations..
A) 2y^210=y.. 2y^2y10=0.. 2y^25y+4y10=0.. y(2y5) + 2(2y5)=0.. (y+2)(2y5)=0.. roots are 5/2 and 2..
B) 2y^210=y.. 2y^2+y10=0.. 2y^2+5y4y10=0.. y(2y+5)  2(2y5)=0.. (y2)(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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If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two [#permalink]
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10 Oct 2017, 01:33
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 \(\frac{9}{2}\).




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