• Jan 10th 2010, 05:19 PM
carla5980
Hello!

I need help with these two problems- they're driving me crazy. Any input would be really appreciated!!

This is the first question:

http://i50.tinypic.com/143erm8.jpg

This is the second question:

http://i46.tinypic.com/315o6mf.jpg

Thanks so so much!
• Jan 10th 2010, 05:24 PM
Prove It
Quote:

Originally Posted by carla5980
Hello!

I need help with these two problems- they're driving me crazy. Any input would be really appreciated!!

This is the first question:

http://i50.tinypic.com/143erm8.jpg

This is the second question:

http://i46.tinypic.com/315o6mf.jpg

Thanks so so much!

1) $\displaystyle x + \frac{1}{x} = 3$

$\displaystyle x^2 + 1 = 3x$

$\displaystyle x^2 - 3x + 1 = 0$

$\displaystyle x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$

$\displaystyle = \frac{3 \pm \sqrt{5}}{2}$.

So $\displaystyle x = \frac{3 - \sqrt{5}}{2}$ or $\displaystyle x = \frac{3 + \sqrt{5}}{2}$.

2) $\displaystyle x^2 - 3x - 5 = 2$

$\displaystyle x^2 - 3x - 7 = 0$

$\displaystyle x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-7)}}{2(1)}$

$\displaystyle = \frac{3 \pm \sqrt{37}}{2}$

So $\displaystyle x = \frac{3 - \sqrt{37}}{2}$ or $\displaystyle x = \frac{3 + \sqrt{37}}{2}$.
• Jan 10th 2010, 05:28 PM
carla5980
Quote:

Originally Posted by Prove It
1) $\displaystyle x + \frac{1}{x} = 3$

$\displaystyle x^2 + 1 = 3x$

$\displaystyle x^2 - 3x + 1 = 0$

$\displaystyle x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$

$\displaystyle = \frac{3 \pm \sqrt{5}}{2}$.

So $\displaystyle x = \frac{3 - \sqrt{5}}{2}$ or $\displaystyle x = \frac{3 + \sqrt{5}}{2}$.

Thanks for the quick response! I understand the first part, but how do I finish it? With the $\displaystyle x^6$ bit?
• Jan 10th 2010, 05:29 PM
Prove It
You know what $\displaystyle x$ is. So you should be able to find $\displaystyle x^6$.
• Jan 10th 2010, 05:39 PM
carla5980
Quote:

Originally Posted by Prove It
You know what $\displaystyle x$ is. So you should be able to find $\displaystyle x^6$.

I'm just not sure, though. This is the part I'm stuck on.
• Jan 10th 2010, 05:44 PM
mr fantastic
Quote:

Originally Posted by carla5980
Hello!

I need help with these two problems- they're driving me crazy. Any input would be really appreciated!!

This is the first question:

http://i50.tinypic.com/143erm8.jpg

[snip]

From the sum of cubes formula: $\displaystyle x^6 + \frac{1}{x^6} = (x^2)^3 + \left(\frac{1}{x^2}\right)^3 = \left( x^2 + \frac{1}{x^2}\right) \left( x^4 - 1 + \frac{1}{x^4}\right)$.

Now note:

1. $\displaystyle \left(x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2} = 9$. Therefore $\displaystyle x^2 + \frac{1}{x^2} = ....$

2. $\displaystyle \left(x^2 + \frac{1}{x^2} \right)^2 = x^4 + 2 + \frac{1}{x^4} = ....$ (use the above value). Therefore $\displaystyle x^4 + \frac{1}{x^4} = ....$
• Jan 10th 2010, 05:51 PM
Sudharaka
Dear carla5980,

Here's another approch,

$\displaystyle x+\frac{1}{x}=3$

$\displaystyle (x+\frac{1}{x})^3=27$

$\displaystyle x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=27$

$\displaystyle x^3+\frac{1}{x^3}+(3\times{3})=27$

$\displaystyle x^3+\frac{1}{x^3}=27-9=18$

$\displaystyle (x^3+\frac{1}{x^3})^2=18^2$

$\displaystyle x^6+\frac{1}{x^6}+2=18^2$

$\displaystyle x^6+\frac{1}{x^6}=324-2=322$

Hope this helps.
• Jan 10th 2010, 05:54 PM
carla5980
Quote:

Originally Posted by mr fantastic
From the sum of cubes formula: $\displaystyle x^6 + \frac{1}{x^6} = (x^2)^3 + \left(\frac{1}{x^2}\right)^3 = \left( x^2 + \frac{1}{x^2}\right) \left( x^4 - 1 + \frac{1}{x^4}\right)$.

Now note:

1. $\displaystyle \left(x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2} = 9$. Therefore $\displaystyle x^2 + \frac{1}{x^2} = ....$

2. $\displaystyle \left(x^2 + \frac{1}{x^2} \right)^2 = x^4 + 2 + \frac{1}{x^4} = ....$ (use the above value). Therefore $\displaystyle x^4 + \frac{1}{x^4} = ....$

I'm really sorry, I totally do not follow this at all. Can someone please do it step by step?
• Jan 10th 2010, 05:58 PM
carla5980
Quote:

Originally Posted by Sudharaka
Dear

Here's another approch,

$\displaystyle x+\frac{1}{x}=3$

$\displaystyle (x+\frac{1}{x})^3=27$

$\displaystyle x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=27$

$\displaystyle x^3+\frac{1}{x^3}+(3\times{3})=27$

$\displaystyle x^3+\frac{1}{x^3}=27-9=18$

$\displaystyle (x^3+\frac{1}{x^3})^2=18^2$

$\displaystyle x^6+\frac{1}{x^6}+2=18^2$

$\displaystyle x^6+\frac{1}{x^6}=324-2=322$

Hope this helps.

THANKS! Thank you so much!!
• Jan 10th 2010, 06:12 PM
mr fantastic
Quote:

Originally Posted by carla5980
I'm really sorry, I totally do not follow this at all. Can someone please do it step by step?

Questions of this sort assume that you know and can do certain things. eg. Do you know the sum of cubes formula?

From 1. it should be obvious that $\displaystyle x^2 + \frac{1}{x^2} = 7$.

Using that result in 2, you get $\displaystyle \left(x^2 + \frac{1}{x^2} \right)^2 = x^4 + 2 + \frac{1}{x^4} = 49$ and so $\displaystyle x^4 + \frac{1}{x^4} = 47$.

Now you substitute 7 and 49 into $\displaystyle \left( x^2 + \frac{1}{x^2}\right) \left( x^4 - 1 + \frac{1}{x^4}\right)$ to get the answer: (7)(46) = 322.

Sorry, but you don't seem to have made much effort in trying to follow the help you were given and filling in the details.