How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} (a-2) = 0 \hspace{5pt} or \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} a = 2$$
$$ Now \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0 $$
I don't know what to do next. Thanks in advance

Since
$$(a-1)x^2+(a-3)x+a+2$$
is a quadratic polynomial, it has, at most, two roots, then $$(a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in $x$ only if $a=2$.
