Question

Interesting problem !!
solve (x + 1/x)^2 - 3/2(x - 1/x) = 4

Answer 1

Inside both brackets 1/x is a fraction...
1 3 1
(x + ---)^2 - ---(x - ----) = 4
x 2 x

I got the values of x from wolfram but I need the actual steps to solve it.....

Answer 2

is this correct equation?\[(x+1/x)^{2}-\frac{3}{2(1-1/x)}=4\]

Answer 3

No, second term is product of
( 1 )
-3/2 and (x - ---)
( x )

Answer 4

so is this correct:\[(x+1/x)^{2}-\frac{3(x-1/x)}{2}=4\]

Answer 5

Second term can be rewritten as

-3(x^2 - 1)
-----------
2x

Answer 6

hmmm - so is this the correct equation?\[(x+1/x)^{2}-\frac{3(x^{2}-1)}{2x}=4\]

Answer 7

yes, the original equation can be put in this form.....
pls keep in mind that in the first bracket it is 1 divided by x added to x...

Answer 8

ok, so I will try and solve the following:\[(\frac{x^{2}+1}{x})^{2}-\frac{3(x^{2}-1)}{2x}=4\]

Answer 9

yes this is also right

Answer 10

\[(\frac{x^{2}+1}{x})^{2}-\frac{3(x^{2}-1)}{2x}=4\]\[\frac{(x^{2}+1)^{2}}{x^{2}}-\frac{3(x^{2}-1)}{2x}=4\]multiply both sides by 2x^2 to get:\[2(x^{2}+1)^{2}-3x(x^{2}-1)=8x^{2}\]expand and simplify to get:\[2x^{4}-3x^{3}-4x^{2}+3x+2=0\]now we just need to factorise this expression:\[(x-2)(x-1)(x+1)(2x+1)=0\]so solutions are x=2, 1, -1 and -1/2

Answer 11

you could also use the general form for solving quartic equations - see: http://www.1728.org/quartic2.htm

Answer 12

Thanks for your help........
Can u pls explain the factorisation part i.e. the step just before the final values of x
how did you factorise......????

Answer 13

I had also got the solutions, after getting the quartic equation, by hit and trial method by substituting values of x......
BUT I need the factorisation method if possible....

Answer 14

basically I tried to split the quartic into two quadratics like so (taking advantage of the knowledge that the first term is 2x^4):\[(2x^{2}+ax+b)(x^{2}+cx+d)\]I then expanded and simplified this to get:\[2x^{4}+(a+2c)x^{3}+(2d+ac+b)x^{2}+(ad+bc)x+bd\]I then compare this to the quartic we are trying to factorise to deduce:\[bd=2\]I then "guessed" that b and d might be integral and chose b=1, d=2. put these back into formulae to get simplified version:\[2x^{4}+(a+2c)x^{3}+(5+ac)x^{2}+(2a+c)x+2\]then compared with quartic to get:\[a+2c=-3\]\[5+ac=-4\]\[2a+c=3\]solving these gives:\[a=3\]\[c=-3\]substitute these values back into initial quartic factorisation to get:\[(2x^{2}+3x+1)(x^{2}-3x+2)\]each quadratic can then be factorised to give the final answer.

Answer 15

Thanks and u get your medal.......☺

I also got two factors from the 2 at the end and got other two by hit & trial........
I'll study yur method at leisure with some more questions......
Thanks a lot for your time and effort, I wish I cud giv u more medals....☺☻☺

Answer 16

Don't worry - I am not here for the medals - I just enjoy teaching (and learning) :-)