what $a$ that makes $(a - \sqrt 2) x^2 + (a - \sqrt 2)x + a - 1$ always below $x$-axis

Question

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?

Answer 1

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 $a<\sqrt{2}$.