$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ 
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.

My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0$$
First condition: If $\forall$ $x>0$, inequality is satisfied, it means $x=0$ must have been the root of the equation $\quad \{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}=0\quad$ at which given inequality would not be satisfied.
Second condition: If one root is real, then other root must be real as complex roots occur as conjugates if all quadratic equation coefficients are real. So $D>=0$
Third condition: $a>0$ as $\forall$ $x>0$, $\quad\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a+\sqrt{2}>0\quad$
First condition:
$$x=0$$
$$a+\sqrt{2}=0$$
$$a=-\sqrt{2}$$
Second condition:
$$D>=0$$
$$4(a^2-2)^2-4(a-\sqrt{2})(a^2+a-3)(a+\sqrt{2})>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a^2-2-a^2-a+3)>=0$$
$$(a-\sqrt{2})(a+\sqrt{2})(a-1)<=0$$
$$a\in \left(-\infty,-\sqrt{2}\right]\quad \cup \quad[1,\sqrt{2}]$$
Third condition:
$$a>0$$
$$(a-\sqrt{2})(a^2+a-3)>0$$
$$(a-\sqrt{2})\left(a-\left(\frac{-1+\sqrt{13}}{2}\right)\right)\left(a-\left(\frac{-1-\sqrt{13}}{2}\right)\right)>0$$
$$a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Taking intersection of all three conditions would give
$$a\in \{-\sqrt{2}\}\quad$$
But answer is $a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$
What am I missing here, I tried to think of it a lot but didn't any breakthroughs. Please help me in this.

For the first condition
($x = 0$),
you want 
$a+\sqrt{2} > 0$
so
$a > -\sqrt{2}$.
