8
$\begingroup$

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?

  • 2
    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.2011-02-05
  • 2
    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: http://en.wikipedia.org/wiki/Derivative2011-02-05

3 Answers 3

15

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

5

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.

4

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.

  • 0
    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..2011-02-20
  • 0
    @Tom: $\rm\ g^2 = 4 + h\ \Rightarrow\ h = g^2-4\:.\:$ That step results from substituting this value for $\rm\:h\:$ into the denominator.2011-02-20