If 4y^4 − 41y^2 + 100 = 0, then what is the sum of the two : GMAT Problem Solving (PS)
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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]

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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. $$\frac{9}{2}$$

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, 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 $$2+\frac{5}{2}=\frac{9}{2}$$.

Hope it 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, 20:08
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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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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]

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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

=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, 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

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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, 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: $$\frac{5}{2} + 2 = \frac{9}{2}$$

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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, 05:18
$$4y^4 − 41y^2 + 100 = 0$$
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, 08:14
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   [#permalink] 04 Mar 2016, 08:14
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