Example OAF.SAE.2

Example OAF.SAE.2 : Suppose $\sqrt {x + 5} = 2x - 5$. Find $x$.
Solution: Examining the equation, we see that it can be solved algebraically by first solving $x+5 = (2x -5)^2 = 4x^2 -20x +25 $ or equivalently $4x^2 - 21x + 20 = (x-4)(4x-5) =0$. This quadratic equation gives the two solutions, $x =4$ and $x= \frac 5 4$. However only $x=4$ satisfies the original equation since $2 *(\frac 5 4) -5 < 0$.

Comment: We can consider the different expressions on the left and right hand sides of the equation as separate functions of $x$, $f$ and $g$, giving $f(x) = \sqrt {x + 5} $ and $g(x) = 2x - 5$. Now the problem can be restated: to find a number $x$ where $f(x) = g(x)$. This problem and its solution can be visualized both on the combined graphs and then mapping diagrams for the functions $f$ and $g$

On the GeoGebra figures move the value of $x$ on the mapping diagram to explore the relation between the values of $f(x)$ (in blue) and $g(x)$ (in red) at $x= \frac 5 4$ and $x=4$..

For the graphs of $f$ and $g$: Find the point(s) on the graph of $f$ and $g$ where the two graphs intersect, determine the first coordinate, $4$, and that is the desired value for $x$.	For the mapping diagrams of $f$ and $g$: Find the point(s) $x$ on the source axis with a single function arrow pointing to $f(x) = g(x)$.

Martin Flashman, 7 December 2013, Created with GeoGebra