Hi folks,

I am getting the wrong range of values to the following question:

calculate the range of x for which:

\[ \left | \frac{2x + 5}{x^2 - 4} \right | \ge \frac{1}{5} \]

the modulus means that both numerator and denominator are positive, therefore:

\[ 2x + 5 \ge \frac{1}{5}(x^2 - 4) \]

$ -x^2 + 10x + 29 \ge 0 $ ....................................(1)

\[ x = \frac{-10 \pm \sqrt{100 + 4.1.29}}{-2} \Rightarrow 5 \pm \sqrt{216} \]

\[ x = 5 \pm 3\sqrt{6} \]

\[ (x - (5 - 3\sqrt{6})) (x - (5 + 3\sqrt{6})) \ge 0 \]

\[ (x - 5 + 3\sqrt{6}) (x - 5 - 3\sqrt{6}) \ge 0 \]

for this inequality to be true:

\[ x - 5 + 3\sqrt{6} \ge 0 \ and \ x - 5 - 3\sqrt{6} \ge 0 \]

\[ x \ge 5 - 3\sqrt{6} \ and \ x \ge 5 + 3\sqrt{6} \]

for both to be true simultaneously x must be $ \ge 5 + 3\sqrt{6} $

or

\[ x - 5 + 3\sqrt{6} \le 0 \ and \ x - 5 - 3\sqrt{6} \le 0 \]

\[ x \le 5 - 3\sqrt{6} \ and \ x \le 5 + 3\sqrt{6} \]

for both to be true simultaneously x must be $ \le 5 - 3\sqrt{6} $

so the range is

\[ 5 + 3\sqrt{6} \le x \le 5 - 3\sqrt{6} \]

but the answer given is

\[ 5 - 3\sqrt{6} \le x \le 5 + 3\sqrt{6} \]

I think the reason for the problem is that (1) can be framed as shown above or written as

$ x^2 -10x -29 \le 0$ ....................................(2)

in both cases the roots are the same $x = 5 \pm 3\sqrt{6}$

so my question to the forum is should I use equation (1) or (2) ? (2) gives the right answer, but I should be able to use either. Is it the roots that should be different? I was a bit unsure when I was taking the negative of $\pm$ !