Matrix solving equation. The number of real solutions of equation $$\begin{vmatrix}x^2-12&-18&-5\\10&x^2+2&1\\-2&12&x^2\end{vmatrix}=0$$
is?
Well I wanted to do something like this:
$$\begin{vmatrix}-12&-18&-5\\10&2&1\\-2&12&0\end{vmatrix}+|x^2{\rm I}|=0$$
And then I got:
$$x^2=704>0$$
Does this prove that it has two real roots?
 A: No. The operation you carried out is not true. Though you got the number of real roots as 2, which is correct, but the real roots are incorrect. The roots of the equation are 2,-2.  
You can either split open one row or a column but not all in the same step.  
For a speedy way to solve it:  
Take $x^2=t$ and then expand. It will take you less than a minute to get the cubic in $t$. Since your original question was just to find the number of roots, the cubic is an increasing function (check its derivative, which being a quadratic has discriminant less than zero) which cuts the real x-axis only once (hence there is one root for $t$) . This root shall be positive, hence you have two real roots (for $x$). Ta-da, it shall take you two minutes max :)
A: You cannot use the ``rule'' $\det (A+B) = \det A + \det B$, as it's just wrong :) Try adding the second row to the first and subtracting the third row from it. Such row operations do not change the determinant.
A: Expanding, $f(x) = -440 + 134 x^2 - 10 x^4 + x^6$. Solving for $y=x^2$, we get two real roots.
Edit: Using Descartes' Rule of signs, it has at most 3 positive (3 or 1), at most 3 negative (3 or 1) roots. 
Letting for $y=x^2$, we have $g(y)=-440 + 134 y - 10 y^2 + y^3$, using Descartes' rule of signs, it has at most 3 positive roots (3 or 1), and no negative roots.
It has at least one real root since the equation is cubic, and Descartes' rule of signs assures that if it is real, then it is positive. That is, there is some $a>0$ such that for $y=x^2=a, g(y)=0$. Hence $x=\pm a$ is a solution to $f(x)$.
That assures $f(x)$ has at least two real roots. (This doesn't assure that number of real roots = 2 which is true).
