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# Let x and y be the solutions of the equation a^2+3a-1=0

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Director
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Let x and y be the solutions of the equation a^2+3a-1=0 [#permalink]

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16 Sep 2017, 09:55
00:00

Difficulty:

75% (hard)

Question Stats:

36% (03:33) correct 64% (02:27) wrong based on 14 sessions

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Let $$x$$ and $$y$$ be the solutions of the equation $$a^2+3a-1=0$$. If $$P_n$$ $$= x^n+y^n$$ for $$n≥0$$, then, for $$n≥2$$, $$P_n =$$

A) $$-3P_{n-1} + P_{n-2}$$

B) $$3P_{n-1} + P_{n-2}$$

C) $$-P_{n-1} + P_{n-2}$$

D) $$P_{n-1} + 3P_{n-2}$$

E) $$P_{n-1} + P_{n-2}$$
[Reveal] Spoiler: OA

Kudos [?]: 247 [0], given: 33

Senior Manager
Joined: 02 Jul 2017
Posts: 269

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Location: India
Let x and y be the solutions of the equation a^2+3a-1=0 [#permalink]

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16 Sep 2017, 11:24
I am not sure about shortcut method. But find below the method i used:

$$a^2+3a−1=0 => x+y = \frac{-3}{1} = -3$$ and $$xy=\frac{-1}{1}=-1$$ ( By sum and product of quadratic equation formula)

$$P_n =x^n+y^n$$ $$n≥0$$

$$P_1 = x+y = -3$$
$$P_2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11$$
$$P_3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36$$

Now with respect to $$P_3$$ =>$$P_2 =P_{n-1}$$ and $$P_1=P_{n-2}$$
but putting values we got $$P_3 =-36 = -3*(11)+(-3) = -3*P_{n-1}+P_{n-2}$$

Last edited by Nikkb on 16 Sep 2017, 13:02, edited 1 time in total.

Kudos [?]: 80 [0], given: 65

Director
Joined: 25 Feb 2013
Posts: 542

Kudos [?]: 247 [0], given: 33

Location: India
Schools: Mannheim"19 (S)
GPA: 3.82
Re: Let x and y be the solutions of the equation a^2+3a-1=0 [#permalink]

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16 Sep 2017, 12:27
Nikkb wrote:
I am not sure about shortcut method. But find below the method i used:

$$a^2+3a−1=0 => x+y = \frac{-3}{1} = -3$$ and $$xy=\frac{-1}{1}=-1$$ ( By sum and product of quadratic equation formula)

$$Pn =x^n+y^n$$ $$n≥0$$

$$p1 = x+y = -3$$
$$p2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11$$
$$p3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36$$

Now for p3 =>$$p2 =p(n-1)$$ and $$p1=p(n-2)$$
but putting values we got $$p3 =-36 = -3*(11)+(-3) = -3*p(n-1)+p(n-1)$$

Hi Nikkb,

We don't need a short-cut method here as the question can be solved in less than 2 minutes. Your approach is correct, however calculations could be made bit simpler to save time.

As you have rightly mentioned: Sum of roots, $$x+y = -3$$ and Product of roots, $$xy=-1$$

Now we need to find $$P_n$$ for $$n≥2$$. This can be done by simply finding $$P_2$$

If you observe all answer choices have only $$P_{n-1}$$ and $$P_{n-2}$$. Hence we need to find $$P_1$$ & $$P_0$$

$$P_0$$ $$= x^0+y^0 = 1+1 =2$$

$$P_1$$ $$= x^1+y^1 = x+y=-3$$

$$P_2$$ $$=x^2+y^2 = (x+y)^2-2xy = (-3)^2-2*-1= -3*-3+2 =$$ $$-3P_1$$ $$+ P_0$$

Or $$P_n$$ $$= −3P_{n−1}$$ $$+ P_{n−2}$$

Hence Option A

Kudos [?]: 247 [0], given: 33

Senior Manager
Joined: 02 Jul 2017
Posts: 269

Kudos [?]: 80 [0], given: 65

Location: India
Let x and y be the solutions of the equation a^2+3a-1=0 [#permalink]

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16 Sep 2017, 12:31
niks18 wrote:
Nikkb wrote:
I am not sure about shortcut method. But find below the method i used:

$$a^2+3a−1=0 => x+y = \frac{-3}{1} = -3$$ and $$xy=\frac{-1}{1}=-1$$ ( By sum and product of quadratic equation formula)

$$Pn =x^n+y^n$$ $$n≥0$$

$$p1 = x+y = -3$$
$$p2 = x^2+y^2 = x^2+y^2+2xy-2xy = (x+y)^2-2xy = (-3)^2-2*(-1) = 11$$
$$p3 =x^3+y^3 =x^3+y^3+3xy(x+y)-3xy(x+y)=(x+y)^3-3xy(x+y) =-36$$

Now for p3 =>$$p2 =p(n-1)$$ and $$p1=p(n-2)$$
but putting values we got $$p3 =-36 = -3*(11)+(-3) = -3*p(n-1)+p(n-1)$$

Hi Nikkb,

We don't need a short-cut method here as the question can be solved in less than 2 minutes. Your approach is correct, however calculations could be made bit simpler to save time.

As you have rightly mentioned: Sum of roots, $$x+y = -3$$ and Product of roots, $$xy=-1$$

Now we need to find $$P_n$$ for $$n≥2$$. This can be done by simply finding $$P_2$$

If you observe all answer choices have only $$P_{n-1}$$ and $$P_{n-2}$$. Hence we need to find $$P_1$$ & $$P_0$$

$$P_0$$ $$= x^0+y^0 = 1+1 =2$$

$$P_1$$ $$= x^1+y^1 = x+y=-3$$

$$P_2$$ $$=x^2+y^2 = (x+y)^2-2xy = (-3)^2-2*-1= -3*-3+2 =$$ $$-3P_1$$ $$+ P_0$$

Or $$P_n$$ $$= −3P_{n−1}$$ $$+ P_{n−2}$$

Hence Option A

niks18
Thanks
Well starting from n=0 makes problem little simple

Added note just in case if someone else have other method than he can share that

[PS: i think i should try to avoid solving questions and posting solutions when i am in between 2 things ]

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Intern
Joined: 27 Nov 2016
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Re: Let x and y be the solutions of the equation a^2+3a-1=0 [#permalink]

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17 Sep 2017, 00:12
Option:A
Put n =2 in the expression.
Since finding P0 and P1 isn't time consuming, we can easily reach to the answer within a minute.

+1 Kudos if it helped.
_________________

+1 Kudos if my post helps in any way

Kudos [?]: 14 [0], given: 9

Re: Let x and y be the solutions of the equation a^2+3a-1=0   [#permalink] 17 Sep 2017, 00:12
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