How do I put $\sqrt{x+1}$ into exponential notation? I think $\sqrt{x+1} = x^{1/2} + 1^{1/2}$. Is this incorrect? Why or why not?
 A: No, it is not correct. $$\sqrt{x+1} = (x+1)^{1/2}\neq x^{1/2} + 1^{1/2}$$
Equality holds only if $x = 0$.
Generally speaking, $(x + y)^p \neq x^p + y^p$, where $p$ is some power. Equality holds if and only if $x=0$ or $y = 0$.
A: Remember the formula for fractional exponents:
$$x^\frac{m}{n} = \sqrt[n]{x^m}$$
It can also be written as:
$$x^\frac{m}{n} = (\sqrt[n]{x})^m$$
$\sqrt{x+1}$ can be rewritten as $\sqrt[2]{(x+1)^1}$
Using our formula, we can say that:
$$\sqrt[2]{(x+1)^1} = (x+1)^\frac{1}{2}$$
Also remember that:
$$x^m + y^m \neq (x+y)^m$$
You should be familiar with the Pythagorean triple $3-4-5$. It can be easily seen that:
$$3^2 + 4^2 \neq (3+4)^2 \neq 7^2$$
$$3^2 + 4^2 = 5^2$$
I will now go back to the original question. You think that:
$$\sqrt{x+1} = x^\frac{1}{2} + 1^\frac{1}{2}$$
Let us expand $x^\frac{1}{2} + 1^\frac{1}{2}$.
$$x^\frac{1}{2} + 1^\frac{1}{2}$$
$$=\sqrt[2]{x^1} + \sqrt[2]{1^1}$$
$$=\sqrt{x} + \sqrt{1}$$
$$=\sqrt{x} + 1$$
$$\sqrt{x} + 1 \neq \sqrt{x+1}$$
So, because $\sqrt{x} + 1 \neq \sqrt{x+1}$, then $\sqrt{x+1} \neq x^\frac{1}{2} + 1^\frac{1}{2}$. The correct answer is:
$$\sqrt{x+1} = (x+1)^\frac{1}{2}$$
Before I forget, always remember the restrictions on $\sqrt{x+1} = (x+1)^\frac{1}{2}$. If $x < -1$. then the statement is false, because you cannot have negative square roots. TECHNICALLY, it is not false because complex numbers cover negative square roots. But for the purposes of this question, UNLESS you have learned about complex numbers and the number $i$, the statement is false if $x < -1$.
RESTRICTIONS ON $\sqrt{x+1} = (x+1)^\frac{1}{2}$:
$x ≥ 1$
$\sqrt{x+1} ≥ 0$
$(x+1)^\frac{1}{2} ≥ 0$
I hope I have enlightened you a bit :)
EDIT: I do not want you to think that the statement $\sqrt{x}+1=\sqrt{x+1}$ is entirely false. There is a value of $x$ that will make the equation true. Let's find it.
$$\sqrt{x}+1=\sqrt{x+1}$$
$$x+2\sqrt{x}+1=x+1$$
$$2\sqrt{x}=0$$
$$4x=0$$
$$x=0$$
Checking for extraneous roots...
$$\sqrt{0}+1=\sqrt{0+1}$$
$$1=\sqrt{1}$$
$$1=1$$
So, $\sqrt{x}+1=\sqrt{x+1}$ when $x=0$.
A: If $\displaystyle\sqrt a+\sqrt b=\sqrt{a+b}$
Squaring either sides 
$\displaystyle\implies (\sqrt a+\sqrt b)^2=a+b\iff 2\sqrt{ab}=0\iff ab=0$
$\implies$ at least one of $a,b$ is zero
So, here $x$ has to be zero
A: $$
9^{1/2} + 16^{1/2} = 3 + 4 = 7 \ne 5 = 25^{1/2} = (9+16)^{1/2}.
$$
So
$$
\left(\frac{9}{16}\right)^{1/2} + 1^{1/2} = \frac 3 4 + 1 = \frac 7 4 \ne \frac 5 4 = \left(\frac{25}{16}\right)^{1/2} = \left(\frac{9}{16}+1\right)^{1/2}
$$
In other words, if $x=\dfrac{9}{16}$ then $x^{1/2}+1^{1/2}$ is not at all the same as $(x+1)^{1/2}$.
