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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, 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

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

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05 Jul 2015, 02:21

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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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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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