Absolute value equations, quadratic equations and functions
Solving quadratic equations with absolute value
Graphing absolute value functions or equations, examples
Graphical interpretation of the definition of the absolute value of a function = f (x) will help us solve an equation with absolute value.
The definition for the absolute value of a function is given by
Thus, for values of x for which f (x) is nonnegative ( > 0 ), the graph of | f (x)| is the same as that of f (x).
For values of x for which f (x) is negative ( < 0 ), the graph of | f (x) | is a reflection of the graph of f (x) on the x-axis. That is, the graph of  = - f (x) is obtained by reflecting the graph of = f (x) across the x-axis.
Hence, the graph of the absolute value of the function = x, i.e., | x | is
Example:  Solve the absolute value equation  | x2 - 3 | = - x + 3.
Solution:  Using the definition of the absolute value of a function we can write,
 if    x2 - 3 > 0   then   | x2 - 3 | = x2 - 3, and given equation   x2 - 3 = - x + 3  or x2 + x - 6 = 0.   By factoring the equation, x2 + 3x - 2 x - 6 = [x(x + 3) - 2(x + 3)] x2 + x - 6 = (x + 3)(x - 2) = 0 so,  x + 3 = 0,  x1 = - 3  and  x - 2 = 0,  x2 = 2. If    x2 - 3 < 0   then   | x2 - 3 | = - (x2 - 3), and given equation   - ( x2 - 3) = - x + 3  or - x2 + x = x(- x + 1) = 0 so,  x3 = 0  and  - x + 1 = 0,   x4 = 1.
Therefore, the solutions of the given equation,  x1 = - 3,  x2 = 2x3 = 0  and  x4 = 1 are the abscissas of the intersection points of,  = | x2 - 3 | and  = - x + 3  functions, as is shown in the above figure.
Example:  Solve the absolute value equation  | 2x2  - x - 3 | = - x + 5.
Solution:  Using the definition of the absolute value of a function we can write,
 if         2x2  - x - 3 > 0 then    | 2 x2  - x - 3 | = 2 x2  - x - 3, and given equation  2 x2  - x - 3 = - x + 5 or       2x2 = 8,     x2 = 4, so,   x1,2 = ± 2,    x1 = - 2  and  x2 = 2. if         2 x2  - x - 3 < 0 then    | 2x2  - x - 3 | = - (2x2  - x - 3), and given equation   - (2 x2  - x - 3) = - x + 5, - 2x2  + 2x - 2 = 0   or   x2  - x + 1 = 0, thus, there are no real solutions in this case.
Therefore, the solutions of the given equation,  x1 = - 2  and  x2 = 2 are the abscissas of the intersection points of,      = | 2 x2  - x - 3 | and  = - x + 5  functions, as is shown in the above figure.
Pre-calculus contents B