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I'm reviewing Algebra in an attempt to review Calculus and came upon a question in the Algebra Diagnostic that asked to rationalize an expression and simplify.

To my understanding, rationalization is the process of rewriting a given expression so that the denominator is non-zero. The equation is, thus:

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

I understand that $h$ can equal zero, and therefore cannot be in the denominator. The answer given is:

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

The denominator in the answer, I believe, is the conjugate of the numerator in the question, but how did they arrive at this state? And in general, is my understanding of rationalization flawed? What is the point and how will that be applied at higher levels of math?

Thank you

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  • $\begingroup$ but how did they arrive at this $\,(a-b)(a+b)=a^2-b^2\,$ with $\,a=\sqrt{4+h}\,$ and $\,b=2\,$. $\endgroup$
    – dxiv
    Feb 7 '18 at 22:48
  • $\begingroup$ can you elaborate on this? i tried to arrive at a solution using the special form you described but couldn't. the way i finally arrived at their solution was simply to multiple by the conjugate as 1 (e.g. conjugate/conjugate) $\endgroup$
    – mburke05
    Feb 7 '18 at 23:06
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    $\begingroup$ $\require{cancel}a^2-b^2=(\bcancel{4}+h)-\bcancel{2^2}=h$, so $(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)=h$ $\iff \frac{\sqrt{4+h} - 2}{h} = \frac{1}{\sqrt{4+h} + 2}$. $\endgroup$
    – dxiv
    Feb 7 '18 at 23:08
  • $\begingroup$ ah, i see now. thank you! $\endgroup$
    – mburke05
    Feb 7 '18 at 23:10
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The expression $\dfrac{\sqrt{4+h} -2} h$ is undefined when $h=0,$ since then both the numerator and the denominator are equal to $0.\vphantom{\dfrac11}$

The expression $\dfrac 1 {\sqrt{4+h} + 2}$ is equal to the expression above whenever $h\ne0,$ but is defined when $h=0,$ and is continuous where $h=0,$ and when $h=0$ it is equal to $\dfrac 1 4.$

One would rationalize the numerator in the first fraction above for the purpose of showing that it approaches $\dfrac 1 4$ as $h$ approaches $0.$

If the denominator of a fraction approaches $0$ while the numerator approaches some nonzero number, then the fraction approaches $\infty,$ in absolute value at least. But if both the numerator and the denominator approach $0,$ then in many cases the fraction approaches some particular number that is not $0$ or $\pm\infty.$ That is important in calculus since derivatives are limits as both the numerator and the denominator of a difference quotient approach $0.$

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  • $\begingroup$ so to be clear; are the expressions you just described different functions given that $h$ is defined in the second and not in the first at $h=0$? even given they are derived from the same expression/function? am i mixing up the definition of expression and function here? $\endgroup$
    – mburke05
    Feb 7 '18 at 23:08
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They rationalized the numerator. This created a factor h in the numerator, allowing the h in the denominator to be divided out. For calculus, this allows the limit as h->0 to be evaluated.

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