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Jim Brier
11-25-2002, 06:13 AM
If the quantity [x + (2/x)]is squared and set equal to 6, what is the quantity [(x cubed) + ((8)/(x cubed))]?

irchans
11-25-2002, 09:47 AM
zero

marbles
11-25-2002, 10:44 AM
Taking the square root of both sides and multiplying through by x, you can solve for x by the quadratic.
X=square root of 6 plus or minus square root of 14 all over 2. Plugging both results into your new equation, the result is approximately 29.39388 either way.

11-25-2002, 01:25 PM
180

11-25-2002, 01:30 PM
(x + 2/x)^2 = 6
(x + 2/x) = + - sqrt(6)
x^2 + 2 = + - (sqrt(6))*x
x^2 + - (sqrt(6))x + 2 = 0

x = ( + - sqrt(6) + - sqrt(6 MINUS 4*1*2) ) / 2

x = (+ - sqrt(6) + - sqrt(-2)) / 2

x is a complex number, however that doesn't mean the problem doesn't have an answer.

marbles
11-25-2002, 01:38 PM
The first step of the quadratic is -b, not +-b. It results in two possible answers for X, but they both give the same answer when plugged into the next equation.

marbles
11-25-2002, 01:41 PM
Oh, I see your point... I was assuming that we are only allowing the + square root of 6. Allowing for negative square root of 6 does make the problem a little more messy.

pudley4
11-25-2002, 01:46 PM
No, if you use the quadratic equation to solve for X, you get an imaginary number, not a real number.

(x+2/x)^2=6
x + 2/x = sqrt(6)
x^2 - sqrt(6)x + 2 = 0

plug into the quadratic equation [-b +/- sqrt(b^2 - 4ac)]/2a

notice b^2 - 4ac = 6 - 8 = -2, so you end up with an imaginary number.

and I see this point has already been made /forums/images/icons/smile.gif

11-25-2002, 01:53 PM
Sorry, I missed the "is squared" in my first reply.
Imagine that!

11-25-2002, 01:54 PM
I am using the notations a^b = a to the b power, and sqrt(c) = the positive square root of c.

Expanding out (x + (2/x))^3, we get x^3 + 6x + 12/x + 8/(x^3).

Rearranging the terms,

(x + (2/x))^3 = x^3 + 8/(x^3) + 6(x + 2/x)

Now, we are given (x + 2/x)^2 = 6. Therefore (x + 2/x) = sqrt(6) or (x + 2/x) = - sqrt(6).

so, we have (sqrt(6))^3 = x^3 + 8/(x^3) + 6*sqrt(6), or
................(-sqrt(6))^3 = x^3 + 8/(x^3) - 6*sqrt(6)

solving each: 6*sqrt(6) = x^3 + 8/(x^3) + 6*sqrt(6)
0 = x^3 + 8/(x^3)

-6*sqrt(6) = x^3 + 8/(x^3) - 6*sqrt(6)
0 = x^3 + 8/(x^3)

In either case, x^3 + 8/(x^3) = 0, which is what we were asked to find. Looks like irchans is correct, it all disappears into nothingness.

pudley4
11-25-2002, 01:56 PM
He's right, it is zero /forums/images/icons/smile.gif

pudley4
11-25-2002, 02:02 PM
The easiest way is to factor the equation:

x^3 + 8x^(-3) becomes

[x+2x^(-1)] * [x^2 - 2 + 4x^(-2)]

If you notice, the second factor can be found from the initial equation:

(x + 2/x)^2 = 6
[x^2 + 4 + 4x^(-2)] = 6
[x^2 - 2 + 4x^(-2)] = 0

Since this factor = 0, the equation equals 0

11-25-2002, 09:08 PM
Here's a complete solution which solves for x:

(x + 2/x)^2 = 6

x^2 + 4 + 4/x^2 = 6

x^4 + 4x^2 + 4 = 6x^2

x^4 - 2x^2 + 4 = 0

x^2 = [2 +/- sqrt(4 - 16)]/2 from quadratic formula

x^2 = [2 +/- j*sqrt(12)]/2 where j = sqrt(-1)

x^2 = 1 +/- j*sqrt(3)

x^2 = 2*exp(+/-j*pi/6) going to polar form

<font color="red">x = +/-sqrt(2)*exp(+/-j*pi/12) =
+/-sqrt(2)*[cos(pi/12) +/- j*sin(pi/12)</font color>

x is complex, not real or imaginary, and has 4 possible values

<font color="red">x^3 - 8/x^3 =
+/-sqrt(8)*exp(+/-j*pi/4) -/+ sqrt(8)*exp(+/-j*pi/4) = 0 </font color>

Jim Brier
11-25-2002, 09:10 PM
The answer is zero. Here is how I did it.

First of all, you should recognize that (x^3) + (8/x^3) is the sum of two cubes. Factoring the sum of two cubes results in: [x+(2/x)][(x^2)-(2)+(4/x^2)].

Now let us work out [x+(2/x)]^2 = 6.

This is [(x^2)+(4)+(4/x^2)] = 6. Then,

[(x^2)-(2)+(4/x^2)] = 0. But this is the second factor when we factored the sum of the cubes above. So, since the second factor is 0, the result must be 0.

BruceZ
11-25-2002, 09:14 PM
x

BruceZ
11-25-2002, 09:41 PM
x^2 = 2*exp(+/-j*pi/<font color="blue">3</font color>) going to polar form

<font color="red">x = +/-sqrt(2)*exp(+/-j*pi/</font color><font color="blue">6</font color><font color="red">) =
+/-sqrt(2)*[cos(pi/6) +/- j*sin(pi/6)]</font color>

= +/-sqrt(6)/2 +/- j*sqrt(2)/2

in areement with Polarbear

x is complex, not real or imaginary, and has 4 possible values

<font color="red">x^3 - 8/x^3 =
+/-sqrt(8)*exp(+/-j*pi/2) -/+ sqrt(8)*exp(+/-j*pi/2) = 0</font color>