341doesnt this belong to IX-Xth sub-forum?
11 Answers
62the observation would have been to divide by x^2
\left(x+2/x \right)^2+8=6(x+2/x)
Lets put x+2/x= t
then you have a quadratic in t which solves to t=2 or 4
now solve... for x+2/x=2 or x+2/x = 4
1i have another one ... a bit lenGthy
(x^2 + 2) ( x^2-6x +2) = - 8 x^2
\textrm{now \: dividing \, by } x^2( x≠0)
\left ( x +\frac{2}{x} \right )\left ( x + \frac{2}{x} -6 \right ) = - 8
let
x +\frac{2}{x} = t
the equation becomes
\left ( t \right )\left ( t-6 \right ) = -8
t^2 - 6t + 8 = 0
\left ( t-4 \right )\left ( t-2 \right ) = 0
\boxed {t=4}\, \, \textrm{ AND} \, \boxed {t=2}
solvin we have
\boxed{x = 1\pm i}
and
\boxed{x = 2\pm \sqrt{2}}
ricky has explained in detail below abt x≠0
62what you have done is not at all different from the original question...
take (x2+2)(x2+2-6x)=-8x2
= (x2+2)(x2+2)-8x2=6x(x2+2)
which isthe original question [3]
1yes sir .........but isn't it a method of rearrangement for sake of convenience of solving basic problem....doesn't it help in simplification.....[60]
62oh! sorry i din notice that you were giving a solution :D sorry i thought it was another problem :D
1I would just like to add that we are allowed to divide by " x 2 " because x = 0 doesn ' t
satisfy the equation .
1thnks ricky ( btw had mentioned it in hidden box)
HAPPY BIRTHDAY @ ricky [96][69][70][76]