I see now that the inequality you referred to was given by JeffM. (The thread is getting too long to read through easily.) He said it was incomplete, not a final answer. He is also wrong about other things, I think (though maybe they are corrected in the additional work he would do).
The main issue is that we can't talk about this without seeing specific work from you! Random questions will not help the discussion along.
I am not sure what was wrong with that answer.
\(\displaystyle x = \dfrac{5}{6} \implies 4x^2 = \dfrac{4 * 25}{36} = \dfrac{25}{9}.\)
\(\displaystyle -\ \dfrac{5}{3} < a < \dfrac{5}{3} \implies 0 \le a^2 < \dfrac{25}{9} \implies -\ \dfrac{25}{9} < -\ a^2.\)
\(\displaystyle \therefore \dfrac{25}{9} - \dfrac{25}{9} < 4x^2 - a^2 \implies 0 < 4x^2 - a^2 \implies ln(4x^2 - a^2) \ \exists.\)
Similarly,
\(\displaystyle x = \dfrac{5}{6} \implies 2x = \dfrac{5}{3}.\)
\(\displaystyle -\ \dfrac{5}{3} < a < \dfrac{5}{3} \implies \dfrac{5}{3} - \dfrac{5}{3} < 2x + a \implies 0 < 2x + a \implies ln(2x + a) \ \exists.\)
If a = 0 and x = 5/6, then
\(\displaystyle \sqrt{5 - 6x} * ln(4x^2 - a^2) = \sqrt{5 - 6x} * ln(2x + a) \implies 0 * ln \left ( \dfrac{25}{9} \right ) = 0 * ln \left ( \dfrac{5}{3} \right ),\)
which is quite obviously true. If a = 1 and x = 5/6, then
\(\displaystyle \sqrt{5 - 6x} * ln(4x^2 - a^2) = \sqrt{5 - 6x} * ln(2x + a) \implies 0 * ln \left ( \dfrac{16}{9} \right ) = 0 * ln \left ( \dfrac{8}{3} \right ),\)
which is also quite obviously true. Moreover, the other restrictions reported by the teacher do not apply in the case of x = 5/6 because as just shown a = 0 works. This suggested to me that the problem was not clearly described, not that the teacher was wrong.