Question

The number of real solutions of the equation |x|2-4|x|+3=0 is
(a)4 (b)2 (c)3 (d)1

Answer 1

Consider the discriminant of quadratic  $x^2-4x+3=0$
$D=b^2-4ac=(-4)^2-4(1)(3)=4>0$
So there are $2$ real solutions to $x^2-4x+3=0$.

Consequently there will be $4$ real solutions to $|x|^2-4|x|+3=0$ as both $x$ and $-x$ satisfy this equation whenever $x$ is a solution to $x^2-4x+3=0$.

Answer 2

x²-4x+3=0
compare with
ax²+bx+c=0
a=1 ,b=-4 and C=3
apply formula
-b+-√b²-4ac    = roots of quadratic equation
    2a
4+-√16-4×1×3       ⇒    4+-√4    ⇒4+-2                                 
        2                            2            2
x=3,1
option c and d are correct