<s>[INST] what $a$ that makes $(a - \sqrt 2) x^2 + (a - \sqrt 2)x + a - 1$ always below $x$-axis it means$(a - \sqrt 2) < 0$ also $D < 0$
so we get $a < \sqrt 2$ and also ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
Solve  ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$
i get $\frac {4 - \sqrt 2}{3} < a < \sqrt 2$
but my answer not in the options. is there something wrong?
 [/INST] Just be careful here:
$$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$$
$$(a - \sqrt 2)[ -3a +4- \sqrt 2] < 0$$
and then:
$$a<\frac{4-\sqrt{2}}{3} \quad \text{or} \quad a>\sqrt{2}$$
and after that you have to take the intesection with $a<\sqrt{2}$.
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