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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