Solving Absolute-Value Equations

l x^2 - 4x - 5 l   = 7    we have two equations here......

The first is :

x^2 - 4x - 5  = 7    subtract 7 from both sides

x^2 - 4x - 12 = 0    factor

(x - 6) ( x + 2)  = 0      and, setting each linear factor  to  0   means that  x= 6 or x = -2

The second equation is :

-(x^2 - 4x  - 5)  = 7       multiply both sides by -1

x^2 - 4x - 5  = -7      add 7 to both sides

x^2 - 4x + 2  = 0     the solution to this is:

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\ {\mathtt{x}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.585\: \!786\: \!437\: \!626\: \!905}}\\ {\mathtt{x}} = {\mathtt{3.414\: \!213\: \!562\: \!373\: \!095}}\\ \end{array} \right\}$$

Thus...we have 4 values that make this true....