| $ (\sqrt{2x+1} + \sqrt{3x+5})^2 $ |
| $ (\sqrt{2x+1} + \sqrt{3x+5})(\sqrt{2x+1} + \sqrt{3x+5}) $ |
| $ (\sqrt{2x+1})(\sqrt{2x+1} + \sqrt{3x+5}) + (\sqrt{3x+5})(\sqrt{2x+1} + \sqrt{3x+5}) $ |
| $ \sqrt{(2x+1)(2x+1)} + \sqrt{(2x+1)(3x+5)} + \sqrt{(3x+5)(2x+1)} + \sqrt{(3x+5)(3x+5)} $ |
If you graph the first 3 expressions, you will see that the graph result is the same. However, as soon as you graph the last expression, you realize that you are working with a different graph result.
What is going on here? The distributive property seems to be right.
The domains are different.
In order to evaluate the expression $(\sqrt{2x + 1} + \sqrt{3x + 5})^2$, we require that $2x + 1 \geq 0 \implies x \geq -1/2$ and that $3x + 5 \geq 0 \implies x \geq -5/3$. For both statements to be true, $x \geq -1/2$. Hence, the domain is $[-1/2, \infty)$.
In order to evaluate the expression $\sqrt{(2x + 1)^2} + 2\sqrt{(2x + 1)(3x + 5)} + \sqrt{(3x + 5)^2}$, we require that $(2x + 1)^2 \geq 0 \implies x \in \mathbb{R}$, that $(2x + 1)(3x + 5) \geq 0 \implies x \leq -5/3$ or $x \geq -1/2$, and that $(3x + 5)^2 \geq 0 \implies x \in \mathbb{R}$. In order for all three statements to be true, $x \leq -5/3$ or $x \geq -1/2$. Hence, the domain is $(-\infty, -5/3] \cup [-1/2, \infty)$.
Consequently, the graph of $y = (\sqrt{2x + 1} + \sqrt{3x + 5})^2$ has a single branch, while the graph of $y = \sqrt{(2x + 1)^2} + 2\sqrt{(2x + 1)(3x + 5)} + \sqrt{(3x + 5)^2}$ has two branches.