Question

Number of real solutions of $(x-2{)}^2+|x^2-4|+\sqrt{x^2-3x+2}=0$

Answer 1

$(x-2{)}^2+|x^2-4|+\sqrt{x^2-3x+2}=0$
As we know that
$(x-2{)}^2\ge 0$
$|x^2-4|\ge 0$
$\sqrt{x^2-3x+2}\ge 0$
But RHS is $0$ it means all three terms will be equal to $0$

$(x-2{)}^2=0$
$x=2\cdots \left(1\right)$

$|x^2-4|=0$
$x^2-4=0$
$x=-2,+2\cdots \left(2\right)$

$\sqrt{x^2-3x+2}=0$
$x^2-3x+2=0$
$(x-2)(x-1)=0$
$x=1,2\cdots \left(3\right)$

Now, all three expressions should be $0$ at the same time so,
By set $\left(1\right)\cap \left(2\right)\cap \left(3\right)$
$x=2$
So, there is only one solution