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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