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

Answer 1

given

$\Rightarrow$ $(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0$

for this equation to be identity in $x$

$\Rightarrow$ $a^2-3a+2=0$ and $a^2-5a+6=0$ and $a^2-4=0$

$\Rightarrow$ $a^2-2a-a+2=0$

$\Rightarrow$ $a(a-2)-1(a-2)=0$

$\Rightarrow$ $(a-2)(a-1)=0$

$\therefore$ $a=2,1$

also

$\Rightarrow$ $a^2-5a+6=0$

$\Rightarrow$ $a^2-3a-2a+6=0$

$\Rightarrow$ $a(a-3)-2(a-3)=0$

$\Rightarrow$ $(a-3)(a-2)=0$

$\Rightarrow$ $a=2,3$

also

$\Rightarrow$ $a^2-4=0$

$\Rightarrow$ $a^2=4$

$\Rightarrow$ $a=\pm 2$

We can observe that for $a=2$

all 3 equations is $0$ and so it is identity in $x$

$\therefore$ For $1$ value of $x$ the above equation is a identity in $x$