$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ 
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.
 A: Why are you taking the intersection of those three conditions? It doesn't make much sense to me. 
For example, the second condition $\Delta\geq 0$ is not necessary. If $a>\sqrt{2}$ then we have $\Delta<0$ and the inequality is satisfied for all real $x$ (not only for $x>0$):
$$\underbrace{(a^2+a-3)(a-\sqrt{2})}_{>0}x^2+2(a^2-2)x+(a+\sqrt{2})>0.$$
Also $a=\sqrt{2}$ works because the inequality boils down to $(\sqrt{2}+\sqrt{2})>0$ which holds.
Hence $[\sqrt{2},+\infty)$ is a subset of the possible values of the parameter $a$.
A: For the first condition
($x = 0$),
you want 
$a+\sqrt{2} > 0$
so
$a > -\sqrt{2}$.
A: Well I got the answer, thanks to @Robert Z
I) First lets see what should be the condition when $a>0 \text{\{a here means coefficient of $x^2$\}}$
So  $$\text{parameter }a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Now there can be two sub-cases:
1) When $D<0$, then at all values of $x$, inequality will hold.
So for $D<0$ and $a>0$, $\text{parameter }a\in \left(-\sqrt{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right)$
2) When $D>=0$, then inequality will hold only when $x=0$ is the root of the given quadratic equation.
So for $a>0$ and $D>=0$ and $\text{$x=0$ is the root}$, we get parameter $a\in \{-\sqrt{2}\}$
II)When $a<0$, as in that case inequality cannot hold till $+\infty$, so no solutions.
III) When $a=0$, then parameter $a$ can be $\{\sqrt{2},\frac{-1+\sqrt{13}}{2},\frac{-1-\sqrt{13}}{2}\}$
If parameter $a=\sqrt{2}$, then $\sqrt{2}+\sqrt{2}>0$ which is always true, so $\sqrt{2}$ is the solution.
If parameter $a=\frac{-1+\sqrt{13}}{2}$, then we get $x<4.5$, so inequality will hold only for $x<4.5 $, but we want inequality to hold $\forall x>0$, so $a=\frac{-1+\sqrt{13}}{2}$ can't be the solution.
If parameter $a=\frac{-1-\sqrt{13}}{2}$, then we get $x>0.133$, so inequality will hold only for $x>0.133$, but we want inequality to hold $\forall x>0$, so $a=\frac{-1-\sqrt{13}}{2}$ can't be the solution.
So taking union of all cases I),II),III), we get $\text {parameter }a\in [-\sqrt{2},\frac{-1+\sqrt{13}}{2})\quad\cup\quad\left[\sqrt{2},\infty\right)$ which is the correct answer.
