# Strange equality

• Oct 6th 2005, 10:34 AM
LoB
Strange equality
x^2 / 3 + 48/x^2 = 10(x/3 - 4/x)
after performing some actions i came to this:
10x(x-2)(x+2) - x^4 - 144 or
x^4 - 10x^3 + 40x +144 = 0

any ideas what to do next, or how to solve it?
• Oct 6th 2005, 01:55 PM
TheBrain
Are you definately sure that is correct and the only information given? I can't see a way of doing that with my knowledge, but there are probably quite a few people around here better than me!
• Oct 6th 2005, 02:12 PM
LoB
yeah, thats it, someone gave an idea of adding -24x^2 to both sides, so some nice grouping could be done... after some trying i got to:
x^2 (x^2 + 1) - 10x(x-2)(x+2) + (12-x)(12+x) = 0
but its still not enough to be able to solve :\
• Oct 7th 2005, 03:48 AM
ticbol
Use synthetic division (trial and error), or, if you know a little bit of Calculus or derivatives, then you can use iteration by Newton's Method.

x^2 / 3 + 48/x^2 = 10(x/3 - 4/x) -------(i)
(x^2 / 3) +(48 / x^2) = (10x / 3) -(40 / x)
Clear the fractions, multiply both sides by 3x^2,
(x^2 *x^2) +(48*3) = (10x *x^2) -(40 *3x)
x^4 +144 = 10x^3 -120x
x^4 -10x^3 +120x +144 = 0 ----------(ii)

So the linear or 1st degree x-term is not +40x, but it is +120x.

You can attack (ii) by synthetic division.
Try any integer-factor of the contant "+144"

Or, try Newton's Method of iteration.
x2 = x1 -[f(x1) / f'(x1)]
x3 = x2 -[f(x2) / f'(x2)]
and so on....
Depending on the precision you want, even only after 3 or 4 iterations you could get one root very close to the correct root of f(x).

In your equation above, it came to
f(x) = x^4 -10x^3 +120x +144
So, f'(x) = 4x^3 -30x^2 +120
Then try a seed root of, say, x1 = 4
etc....

You should get x=6 as one real root in whatever way you may use.
Check x=6 in (i),
x^2 / 3 + 48/x^2 = 10(x/3 - 4/x) -------(i)
6^2 / 3 +48/ 6^2 =? 10(6/3 -4/6)
36/3 +48/36 =? 10(2 -2/3)
12 +4/3 =? 20 -20/3
Multiply both sides by 3,
36 +4 =? 60 -20
40 =? 40
Yes, so, Ok.