# Math Help - Algebra equations help, please?

1. ## Algebra equations help, please?

Hello!

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

This is the first question:

This is the second question:

Thanks so so much!

2. 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:

This is the second question:

Thanks so so much!
1) $x + \frac{1}{x} = 3$

$x^2 + 1 = 3x$

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

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

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

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

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

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

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

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

So $x = \frac{3 - \sqrt{37}}{2}$ or $x = \frac{3 + \sqrt{37}}{2}$.

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

$x^2 + 1 = 3x$

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

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

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

So $x = \frac{3 - \sqrt{5}}{2}$ or $x = \frac{3 + \sqrt{5}}{2}$.
Thanks for the quick response! I understand the first part, but how do I finish it? With the $x^6$ bit?

4. You know what $x$ is. So you should be able to find $x^6$.

5. Originally Posted by Prove It
You know what $x$ is. So you should be able to find $x^6$.
I'm just not sure, though. This is the part I'm stuck on.

6. 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:

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

Now note:

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

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

7. Dear carla5980,

Here's another approch,

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

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

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

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

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

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

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

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

Hope this helps.

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

Now note:

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

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

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

9. Originally Posted by Sudharaka
Dear

Here's another approch,

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

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

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

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

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

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

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

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

Hope this helps.
THANKS! Thank you so much!!

10. 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 $x^2 + \frac{1}{x^2} = 7$.

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

Now you substitute 7 and 49 into $\left( x^2 + \frac{1}{x^2}\right) \left( x^4 - 1 + \frac{1}{x^4}\right)$ 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.