How do you solve this quadratic equation?

Question

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

Answer 1

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 $a=2$.

Answer 2

HINT:

If $Ax^2+Bx+C=ax^2+bx+c$ is an identity

$A=a,B=b,C=c$