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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
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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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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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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