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# Algebra Basics.

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Manager
Joined: 19 Jun 2009
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17 Jul 2009, 02:20
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I knew trick to solve these problems, but I forgot.

1. $$x^8 + \frac{1}{x^8} = 2207$$. Find value of $$x + \frac{1}{x}$$ .

2. $$x + \frac{1}{x} = 2$$. Find value of $$x^9 + \frac{1}{x^9}$$ .

I tried to find similar problems on forum, but could not.
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Re: Algebra Basics. [#permalink]

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17 Jul 2009, 05:14
alpeshvc wrote:
I knew trick to solve these problems, but I forgot.

1. $$x^8 + \frac{1}{x^8} = 2207$$. Find value of $$x + \frac{1}{x}$$ .

2. $$x + \frac{1}{x} = 2$$. Find value of $$x^9 + \frac{1}{x^9}$$ .

I tried to find similar problems on forum, but could not.

i think you can write it as X^8 + X^(-8) = 2207 and then just take 2207^-8 to get it to the desired state? The reverse for the 2nd one where you just do 2^9 to get from x^1 + x^(-1) to x^9 + x^-9. it is a bit rusty on me too....
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Manager
Joined: 19 Jun 2009
Posts: 141
Location: India
Concentration: Strategy, General Management
GMAT 1: 650 Q51 V27
GMAT 2: 710 Q48 V39
GPA: 3.2
WE: Engineering (Computer Hardware)
Followers: 5

Kudos [?]: 18 [0], given: 21

Re: Algebra Basics. [#permalink]

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17 Jul 2009, 19:23
I was able to solve second Q. We just simply take cube & middle terms are replaceable.

In first Q, 2207 is a prime number. So, we can't take it's square root or anything like that. I am still not able to find solution for that. I just know the answer for that problem - 3.
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Re: Algebra Basics. [#permalink]

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21 Dec 2009, 15:30
I think I can help

If we suppose that $$x+\frac{1}{x}$$=b

then we can find:
$$(x+\frac{1}{x})^2=b^2$$
$$x^2+2+\frac{1}{x^2}=b^2$$
$$x^2+\frac{1}{x^2}=b^2-2$$
$$(x^2+\frac{1}{x^2})^2=(b^2-2)^2$$
$$x^4+2+\frac{1}{x^4}=(b^2-2)^2$$
$$x^4+\frac{1}{x^4}=(b^2-2)^2-2$$
$$(x^4+\frac{1}{x^4})^2=((b^2-2)^2-2)^2$$
$$x^8+2+\frac{1}{x^8}=((b^2-2)^2-2)^2$$
$$x^8+\frac{1}{x^8}=((b^2-2)^2-2)^2-2$$

and now to find $$x+\frac{1}{x}$$ we will solve $$((b^2-2)^2-2)^2-2=2207$$

$$((b^2-2)^2-2)^2=2209=47^2$$
$$(b^2-2)^2-2=47$$
$$(b^2-2)^2=49$$
$$b^2-2=7$$
$$b^2=9$$
$$b=3$$ or $$b=-3$$

and it means that $$x+\frac{1}{x}=3$$ or $$x+\frac{1}{x}=-3$$.

To solve second problem it is enough to exponentiate the equation twice:

$$x+\frac{1}{x}=2$$
$$(x+\frac{1}{x})^3=2^3$$
$$x^3+3x+\frac{3}{x}+\frac{1}{x^3}=8$$
$$x^3+\frac{1}{x^3}=8-(3x+\frac{3}{x})$$
$$x^3+\frac{1}{x^3}=8-3(x+\frac{1}{x})$$
$$x^3+\frac{1}{x^3}=8-6$$
$$x^3+\frac{1}{x^3}=2$$

$$(x^3+\frac{1}{x^3})^3=2^3$$
$$x^9+3x^3+\frac{3}{x^3}+\frac{1}{x^9}=2^3$$
$$x^9+\frac{1}{x^9}=8-(3x^3+\frac{3}{x^3})$$
$$x^9+\frac{1}{x^9}=8-3(x^3+\frac{1}{x^3})$$
$$x^9+\frac{1}{x^9}=8-6$$
$$x^9+\frac{1}{x^9}=2$$

that's all
Re: Algebra Basics.   [#permalink] 21 Dec 2009, 15:30
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# Algebra Basics.

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