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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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Question Stats: 65% (02:12) correct 35% (02:13) wrong based on 135 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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if p^4-49p^2+600=0, what is the sum of the two greatest possible value  [#permalink]

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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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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 from #1 and #2 above.
Sum of positive roots $$(5)$$ and $$(+ 2\sqrt{6}) =$$
$$5 + 2\sqrt{6}$$

* 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. (We can't make them sum to units digit 9, but try anyway to get a sense of the numbers.)
(2*2*5)= 20, and (20*20)= 400, so try 20 as one factor. Other factor must be (2*3*5)= 30.
(20*30) = 600, but 20+30 = 50. No good. We must be close. Try 25.(25*24)= 600. Done.

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

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_________________ Re: if p^4-49p^2+600=0, what is the sum of the two greatest possible value   [#permalink] 14 May 2019, 00:35
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