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if p^4-49p^2+600=0, what is the sum of the two greatest possible value

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if p^4-49p^2+600=0, what is the sum of the two greatest possible value [#permalink]

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New post 19 Mar 2018, 22:28
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Difficulty:

  55% (hard)

Question Stats:

59% (03:08) correct 41% (01:57) wrong based on 76 sessions

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if \(p^4-49p^2+600=0\), what is the sum of the two greatest possible values of p?
A. 49
B. 7
C. \(5+2\sqrt{6}\)
D.\(4\sqrt{6}\)
E.0
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Joined: 13 Apr 2013
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Concentration: International Business, Operations
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if p^4-49p^2+600=0, what is the sum of the two greatest possible value [#permalink]

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New post 19 Mar 2018, 22:54
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rishabhmishra wrote:
if \(p^4-49p^2+600=0\), what is the sum of the two greatest possible values of p?
A. 49
B. 7
C. \(5+2\sqrt{6}\)
D.\(4\sqrt{6}\)
E.0


\(p^4-49p^2+600=0\)

Let \(p^2 = p\) ------- (I)

Then we have,

\(p^2 - 49p + 600 = 0\)

\(p^2 - 25p - 24p + 600 = 0\)

\(p(p - 25) - 24(p - 25) = 0\)

\((p - 25)(p - 24) = 0\)

\(p - 25 = 0, p = 25\)

\(p - 24 = 0, p = 24\)

We know \(p = p^2\) from (I)

Replace p with p^2

\(p^2 = 25\)

\(p = +/- 5\), but we need the sum of greatest values we must take positive value of "p".

&

\(p^2 = 24\),

\(p = +/-2\sqrt{6}\), but we need the sum of greatest values we must take positive value of "p"

Therefore, the sum of the two greatest possible values of \(p = 5 + 2\sqrt{6}\)

(C)
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if p^4-49p^2+600=0, what is the sum of the two greatest possible value [#permalink]

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New post 20 Mar 2018, 14:38
rishabhmishra wrote:
if \(p^4-49p^2+600=0\), what is the sum of the two greatest possible values of p?
A. 49
B. 7
C. \(5+2\sqrt{6}\)
D.\(4\sqrt{6}\)
E.0

QZ 's method is elegant.Another way:

\(p^4-49p^2+600=0\)

\((p^2 - 24)(p^2 - 25)\)

\(p^2 = 24\) and \(p^2 = 25\)


Those are the two equations for the roots of the quadratic equation.*
1) \(\sqrt{p^2}=\sqrt{25}\)
\(p= \sqrt{25}= (±) 5\)

2) \(\sqrt{p^2}=\sqrt{24}\)
\(\sqrt{p^2}=\sqrt{24}\)
\(p=\sqrt{4*6}\)
\(p = (±) 2\sqrt{6}\)


The sum of the two greatest possible values of p will be the two positive roots. Sum =
\(5 + 2\sqrt{6}\)

Answer C

* To find factors of 600 that sum to 49, try prime factorization. \(600= 2*2*2*3*5*5\)
One factor must end in 0, 2, or 5 (units digit of 600=0).
Group prime factors.
If 0, the other factor doesn't matter.
(20*30) = 600, but 20+30 = 50. No good. We must be close. Try 25.(25*24)= 600. Done.

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if p^4-49p^2+600=0, what is the sum of the two greatest possible value   [#permalink] 20 Mar 2018, 14:38
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