help needed very important If-then

D

Demio

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May 8, 2015
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I need help with few sums !

if a+1/a= √ 2
then a^4+1/a^4 = ?
please do reply !

and if x^2+1/x^2=83
then x-1/x= ? :confused::confused:
 
Denis said:
To start, those are not "sums".

In the 1st one (left side), I think you meant (a+1) / a;
if you don't understand what I'm saying,
then you need classroom help.
Click to expand...
It's just a normal IF and THEN problem

the problem is actually :

\(\displaystyle \mbox{If }\, a\, +\, \dfrac{1}{a}\, =\, \sqrt{2\,}\, \mbox{ then find the value of }\, a^4\, +\, \dfrac{1}{a^4}\)
 
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if you can solve it ..please do it with steps ...
steps are important ..those are what i want ..
 
Demio said:
It's just a normal IF and THEN problem

the problem is actually :

\(\displaystyle \mbox{If }\, a\, +\, \dfrac{1}{a}\, =\, \sqrt{2\,}\, \mbox{ then find the value of }\, a^4\, +\, \dfrac{1}{a^4}\)
Click to expand...

\(\displaystyle a + \dfrac{1}{a} \ = \ \sqrt{2}\)

What do you get if you square both sides?
 
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Demio said:
I need help with few sums !

if a+1/a= √ 2
then a^4+1/a^4 = ?
please do reply !

and if x^2+1/x^2=83
then x-1/x= ? :confused::confused:
Click to expand...
For the first problem, multiple through by a to get a quadratic equation in a, solve for a and do the a4. Or, you could just square a+1/a, rearrange and square again.

For the second, do the same sort of thing to get a quadratic formula in x2. Or, again, just square (x-1/x) and see what you get.
 
Demio said:
I need help with few sums !

if a+1/a= √ 2
then a^4+1/a^4 = ?
please do reply !

and if x^2+1/x^2=83
then x-1/x= ? :confused::confused:
Click to expand...
I do not get why you refuse to follow instructions and square both sides but rather place more posts. I am upset with this post and will just solve the problem. I hope that nobody minds my doing so.

a+1/a= √ 2
(a+1/a)^2= (√ 2)^2
a^2 + 2 + 1/a^2 = 2
So a^2 + 1/a^2 = 0
As I said, square again
(a^2 + 1/a^2)^2 = 0^2
a^4 + 2 + 1/a^4 =0
You are on your own to figure out what a^4 + 1/a^4
Next time follow instructions
 
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Denis said:
Well, if that's the case, then no real solutions.
Rearrange as a quadratic:
a^2 - SQRT(2)a + 1 = 0
solve to get:
a = [SQRT(2) +- SQRT(-2)] / 2
That's a real mess!

I still think the problem is (a+1) / a = SQRT(2)
Then:
a + 1 = aSQRT(2)
aSQRT(2) - a = 1
a = 1 / [SQRT(2) - 1]
Click to expand...
Dennis, in the end you might be right but look at how nice it worked out for me using a + 1/a =sqrt(2)
 
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Otis said:
Incorrect. The solution is an Integer.
Click to expand...

Denis is correct in that there are no real solutions for a. The values for a are the fourth roots of -1; \(\displaystyle \pm\, e^{\pm\, \frac{i\, \pi}{4}}\)
 
Denis said:
...Quit drinking :twisted:
Click to expand...
Ain't going to happen, tonight's beer and Star Gate night.

Anyway I was talking about the square: Starting from Jomo's
a2 + 1/a2 = 0
we have
a4 + 1 = 0
or
a4 = -1
which, BTW, says
a4 + 1/a4 = -2
or, another way,
(a4 + 1)2 = 0
or 02 = 0
but then I think I already knew that.
 
Denis said:
I'm evidently missing something...what zit thou meanest, Mark?
Click to expand...

Seems to me y'all are talking about two different things. If
a + 1/a = \(\displaystyle \sqrt{2}\)
then
a4 + 1/a4 = c
where c is an integer. The constant a is not a real number.
 
Denis said:
OK OK...I was only looking at solving a + 1/a = SQRT(2);
disregarded the remaining part...my fault...

multiply by a:
a^2 + 1 = aSQRT(2)
square both sides:
a^4 + 2a^2 + 1 = 2a^2
so:
a^4 = -1

a^4 + 1/a^4 = -1 + 1/(-1) = -2

Soooo....what was the big deal :confused:
Click to expand...
Maybe it was not a big deal but you did answer the wrong question.
What has been amusing me watching this thread is that such a simple problem has been getting posts for days.
Take my advise everyone and let this be the last post. There are more important things to do! Like everyday life and helping students.
 
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