(X^2 - 8)/(|X|-2) < 1.
how solve? thx!
Hi,
Let y = |x| and recall that |x|^2 = x^2.
Then our original inequality becomes (y^2 - 8) / (y - 2) < 1, where y is never 2 (it is a good practice to exclude inadmissible solutions at the beginning).
Multiplying both sides by (y - 2)^2, we obtain (y^2 - 8)(y - 2) < (y - 2)^2.
Factoring out (y - 2) and simplifying further, we obtain (y - 2)(y - 3)(y + 2) < 0, from which we arrive at
y < -2 or 2 < y < 3.
Since y >= 0 (because |x| >= 0), y < -2 is inadmissible.
So we only have 2 < |x| < 3, which means that |x| > 2 and |x| < 3.
We will need to take the intersection of {x such that x < -2 or x > 2} and {x such that -3 < x < 3}.
This leads to the solution -3 < x < -2 or 2 < x < 3.
Thanks!
P.S. Usually, Cambridge would start with (y^2 - 8) / (y - 2) < 1 to be solved and then ask the student to deduce the solution of (x^2 - 8) / (|x| - 2) < 1. It is common to see schools making the question slightly more challenging to students. Thanks!
Cheers,
Wen Shih
nice! thx!