# How to simplify $\frac{\sqrt{4+h}-2}{h}$

The following expression:

$$\frac{\sqrt{4+h}-2}{h}$$

should be simplified to:

$$\frac{1}{\sqrt{4+h}+2}$$

(even if I don't agree that this second is more simple than the first).

The problem is that I have no idea of the first step to simplify that.. any help?

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The magic words are "multiply by the conjugate". For what it is worth, I would actually prefer the former form to the latter. I find it is easier to keep radicals out of denominators, so I would call the former the simplification, not the latter. – Arturo Magidin Feb 5 '11 at 21:25
I prefer the latter because the removable singularity is removed. But I also am prone to say $\sin\frac{\pi}{4}=\frac{1}{\sqrt 2}$ rather than $\frac{\sqrt 2}{2}$. @Tom: An important use of such "simplification" is that the latter expression indicates how the original expression can be continuously extended to $h=0$. This allows you to determine that the slope of the tangent line to the curve $y=\sqrt x$ at the point $(4,2)$ is $\frac{1}{4}$. If you haven't already learned derivatives, these ideas are explained in the following article: en.wikipedia.org/wiki/Derivative – Jonas Meyer Feb 5 '11 at 21:59

If you multiply both the top and the bottom by $\sqrt{4+h}+2$, you get $\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}$, which simplifies to $\frac{h}{h(\sqrt{4+h}+2)}$. Then, divide both by $h$ (assuming $h\neq 0$), and you get $\frac{1}{\sqrt{4+h}+2}$.

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It is really simple. Let us just do what is most intuitive, multiply numerator and denominator with what you want to have in denominator. You get: $$\frac{(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)}{h(\sqrt{4+h}+2)}$$ Then observe the numerator has a difference of squares. Multiply the numerator easily using that and then your left with $$\frac{h}{h(\sqrt{4+h}+2)}$$ Just assume $h \neq 0$ and get "rid" of it.

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HINT $\rm\displaystyle\quad\quad g^2 = 4+h\ \ \Rightarrow\ \ \frac{g-2}h\ =\ \frac{g-2}{g^2-4}\ =\ \frac{1}{g+2}$

Usually the "simplification" is the opposite inference - known as rationalizing the denominator.

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Is that step $\frac{g-2}{g²-4}$ correct? If you multiply g+2 up and down, you get $\frac{g²-4}{h(g+2)}$, and you can't go on from here.. – Tom Brito Feb 20 '11 at 19:27
@Tom: $\rm\ g^2 = 4 + h\ \Rightarrow\ h = g^2-4\:.\:$ That step results from substituting this value for $\rm\:h\:$ into the denominator. – Bill Dubuque Feb 20 '11 at 19:42

So its not about simplification. You just want to show they are equal. What you do is put an equal sign between them, and cancel everything you can, if you get 1=1 or similar you are done.

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I have downvoted, because I think this answer is misleading. You have to be very careful when trying to prove an identity not to start with the equation you want to prove and arrive at another equation via potentially irreversible steps. It is best to work with just one side of the alleged identity at a time. – Jonas Meyer Feb 5 '11 at 21:04
@Jonas What is an irreversible step ? Do you not have to be equally careful when when working with one side? – TROLLKILLER Feb 5 '11 at 21:07
Let us prove that $-1$ can be simplified to $1$. So I'll put an equal sign between them, $-1=1$. Now if I square both sides, $1=1$. This is true, so $-1=1$. Don't get me wrong, I'm not saying that you are advocating such an illogical step, but starting by assuming (at least in appearance) what you are supposed to prove has the potential to lead to errors. – Jonas Meyer Feb 5 '11 at 21:11
You have a good point about being careful with one side, but there the rule is that you never even change what the expression is. You can multiply by $1$, add $0$, factor, cancel, etc., but you typically don't do anything that actually changes the value. (Of course, this warning is only for beginners, not those who have enough experience to recognize that there are many valid ways to prove an identity.) – Jonas Meyer Feb 5 '11 at 21:14
user1708 is correct, starting with an identity and following reverse implications is an entirely correct way to reason. It's just that reverse implication only follows from application of a function if that function is invertable. Multiplying by zero obviously isn't invertable. Just because there is an incorrect way to do something doesn't mean there isn't a correct way to do it. Many identities are made incredibly difficult if you try to solve them by a series of object equivalences. – DanielV Mar 29 '14 at 22:40