SOLUTION: For the function f(x) = |x2 - 3| - 2x, find the values of x such that f(x) = 0

You have an equation

    |x^2 -3| - 2x = 0.       (1)


The major idea, strategy and methodology is to divide the entire number line in parts, where the function 
under the absolute value remains positive or negative and then, based on the definition of the absolute value, take off 
the absolute value signs and simplify the given equation in this way.


Then we solve the simplified equation associated with each partial interval, find the root / (the roots) and check
(notice, it is VERY important part of the solution) if the found root belongs to the current partial interval.


Now let's look how it works for the given equation.


1.  The quadratic binomial  (x^2 - 3) is negative over the interval   < x < . 

    Therefore, in this interval,  |x^2-3| = -(x^2-3), and the original equation take the form

        -(x^2 - 3) - 2x = 0.      (2)

    Simplify and solve it

        x^2 + 2x - 3 = 0       (2')   ====>  

         =  =  = .

        So, the equation (2')  has two roots   =  = 1  and   =  = -3.

        Of these two roots, the root  = 1 lies in the interval  (,) 

            THEREFORE, it is also the root of the original equation (1).


        From the other side, the root  = -3 doe not belong to the interval  (,) 

            THEREFORE, it does not follow from our consideration that the root of  = -3 is the root of the original equation (1).


         At this point, we completed our consideration of the interval  (,).



2.  The quadratic binomial  (x^2 - 3) is positive outside the interval   < x < .

    This outside part, as you understand, consists of two semi-infinite intervals

    (,)  and  (,).


    Therefore, in this outside part,  |x^2-3| = x^2-3 by the definition, and the original equation take the form

        x^2 - 3 - 2x = 0.      (3)

     (Notice that it is DIFFERENT from equation (2)).  Simplify and solve it

        x^2 - 2x - 3 = 0          (3')   ====>  

         =  =  = .

        So, the equation (3')  has two roots   =  = 3  and   =  = -1.



        Of these two roots, the root  = 3 lies out the interval  (,) 

            THEREFORE, it is also the root of the original equation (1).


        From the other side, the root  = -1 does belong to the interval  (,) 

            THEREFORE, it does not follow from our consideration that the root of  = -1 is the root of the original equation (1).


         Combining our results from n.1 and n.2, we conclude that the roots of the original equation (1) are  1  and  3.



4.  Now it is good time to look into the plot and to see the big picture visually.



    


    Plot y = |x^2-3| - 2x


What you see in the figure, completely confirms my answer.