Absolute value equations, quadratic equations and functions
Solving quadratic equations with absolute value
Graphing absolute value functions or equations, examples
Quadratic equations with absolute value
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