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If \(\displaystyle x^2+\frac{1}{x^2}=7\), find all possible values of \(\displaystyle x^5+\frac{1}{x^5}\).
If \(\displaystyle x^2+\frac{1}{x^2}=7\), find all possible values of \(\displaystyle x^5+\frac{1}{x^5}\).
Hi Prove It,[tex]\displaystyle \begin{align*} x^2 + \frac{1}{x^2} &= 7 \\ x^2 \left( x^2 + \frac{1}{x^2} \right) &= 7x^2 \\ x^4 + 1 &= 7x^2 \\ x^4 - 7x^2 + 1 &= 0 \\ X^2 - 7X + 1 &= 0 \textrm{ if }X = x^2 \\ X^2 - 7X + \left( -\frac{7}{2} \right) ^2 - \left( -\frac{7}{2} \right) ^2 + 1 &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{49}{4} + \frac{4}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 - \frac{45}{4} &= 0 \\ \left( X - \frac{7}{2} \right) ^2 &= \frac{45}{4} \\ X - \frac{7}{2} &= \pm \frac{ 3\sqrt{5}}{2} \\ X &= \frac{7 \pm 3\sqrt{5}}{2} \\ x^2 &= \frac{7 \pm 3\sqrt{5}}{2} \\ x &= \pm \sqrt{ \frac{7 \pm 3\sqrt{5}}{2} } \\ x &= \pm \frac{\sqrt{14 \pm 6\sqrt{5}}}{2} \end{align*} [/tex]
so
[tex]\displaystyle \begin{align*} x^5 + \frac{1}{x^5} &= \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 + \frac{1}{ \left( \pm \frac{ \sqrt{ 14 \pm 6\sqrt{5} } }{2} \right) ^5 } \end{align*}[/tex]
which probably simplifies quite a lot
Well done, MarkFL! Your answer is correct and your method of approaching it is different from mine as I used trigonometric method to solve it and it's more tedious and lousy than yours!My solution:
\(\displaystyle \left(x+\frac{1}{x} \right)^2=x^2+2+\frac{1}{x^2}=9\)
Hence:
\(\displaystyle x+\frac{1}{x}=\pm3\)
Also, we find:
\(\displaystyle \left(x+\frac{1}{x} \right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x} \right)\)
\(\displaystyle \pm27=x^3+\frac{1}{x^3}\pm9\)
\(\displaystyle x^3+\frac{1}{x^3}=\pm18\)
And finally:
\(\displaystyle \left(x+\frac{1}{x} \right)^5=x^5+\frac{1}{x^5}+5\left(x^3+\frac{1}{x^3} \right)+10\left(x+\frac{1}{x} \right)\)
Hence:
\(\displaystyle x^5+\frac{1}{x^5}=\pm(243-5\cdot18-10\cdot3)=\pm123\)
Hi kaliprasad,Good solution by MarkFL
Here is another
Let f(n) = x^n + 1/x^n = x^n + x^(-n) note it should be strictly f(n,x) but I have taken x implicitly)
Then f(0) = 2
Now
f(n)f(m) = (x^n + x^(-n)) ( x^m + x^(-m) = x^(n+m) + x^(n-m) + x^-(n-m) + x^(-(n+m))
= x^(n+m) + x^(-(n+m)) + x^(n-m) + x^-(n-m)
= f(n+m) + f(n-m)
so we got f(n)f(m) = f(n+m) + f(n-m)
n= 1,m= 1
give f(2) + f(0) = f(1)^2 = 9 so f(1) = 3 or – 3
for f(1) = 3
now n = 2, m= 1 we get
f(2)f(1) = f(3) + f(1) or f(3) = f(2) f(1) – f(1) = 7 * 3 – 3 = 18
n =3 , m= 2 give
f(3)f(2) = f(5) + f(1) or f(5) = f(3) f(2) – f(1) = 18 * 7 – 3 = 126 -3 = 123
for f(1) = - 3 we can compute similarly
My solution:I still would like to see it!![]()
Thank you for your kind words, Mark! I agree with all that you said and you're always a joy to read!I just want to say it speaks highly of your character to be willing to post your solution, even if you feel it is not as straightforward as that given by another. But, I think others benefit by seeing different approaches, so you have done a good thing for MHB.
I know I have posted solutions to problems, only to have someone else come along and post a much more direct and simple solution, and the feeling of "D'oh! Why didn't I think of that?" can momentarily be very powerful!
But, in the end, we can console ourselves with the knowledge that at least we were able to solve the problem, and we can learn from the solutions given by others.
Thank you for sharing your solution.![]()
Hi Dreamweaver,Extremely well said, Mark^^
For me personally, when an problem I find interesting gets posted on here - which thankfully happens a lot - I enjoy looking at all of the different solutions. And sometimes, like in this case, I find the less obvious / more circumspect ones the most interesting...
Well done Anemone! And thanks for sharing!!![]()