Rationalize the numerator of each of the following expressions

Question

√a - 2 / a -4

the answer is supposed to be 1 / √a + 2

Please show me the steps of how you got this answer please

Answer 1

First you MUST use parentheses. $1/\sqrt{a} + 2 $ means first take the square root of a, take the reciprocal of that, and then add 2 to the result. For example if $a = 9$, then $1/\sqrt{9} + 2 = 1/3 + 2 = 2 \ \ 1/3$ or $7/3$

This is not at all the same as $1/ (\sqrt{a} + 2)$ which means take the square root of a, add 2, then take the reciprocal of the result. Here if $a = 9$, then $1/ (\sqrt{9} + 2) = 1/(3+2) = 1/5$.

If you don't pay attention to such things, your answer and the answers teachers are expecting will not match up and you will get more and more confused and start losing a lot of points.

The other person who answered here knows enough about math to figure out what you intended. On tests and more advanced work you don't want to bet your results on mind-reading and kindness. Make a point of being clear.

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It is always good when you see a pattern and have a quick way of figuring out a problem. But it is important to have a system when luck doesn't work out.

I'm going to explain the reasoning process in detail here. Take the time and learn this as it has many surprising uses.

Here, the hint is right in the title of your question. You were asked to rationalize the numerator. To rationalize a real (or complex) number including square roots, you want to eliminate square roots -- usually from the denominator but sometimes (as in this question) from the numerator. There are two fairly simple cases:

(a) one term, as for example $1/\sqrt{3}$ To rationalize the denominator (so you aren't trying to divide by an infinitely long decimal) you multiply by a fancy form of 1 using the same square root you already have, divided by itself: $1 = \sqrt{3} / \sqrt{3} $

$\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * 1 = \frac{1}{\sqrt{3}} *\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

The original fraction $1/\sqrt{3} = 1/1.73205081...$, a horrible division to do, while the new format which is EQUAL, just looks different, is $\sqrt{3}/3 = 1.73205081.../3 = .57735027...$, much easier to deal with.

(b) two terms, as in for example $1/(\sqrt{5} - 1)$ Here we use the properties of the difference of two squares. We know that for any $p$ and $q$, $(p + q)(p - q) = p^2 - q^2 $ So if we have a $(p+q)$ we multiply by $(p-q)$ and if we have a $(p-q)$ we multiply by $(p + q)$. This form, the same binomial with the opposite sign in the middle, is called the conjugate of the original binomial. This concept of the conjugate becomes important and is used in many places later, so learn and remember it.

Again we multiply our original fraction by a fancy form of 1, the conjugate over itself.

In my example problem,

$\frac{1}{(\sqrt{5} - 1)} = \frac{1}{(\sqrt{5} - 1)} * 1 = \frac{1}{(\sqrt{5} - 1)} * \frac{(\sqrt{5} + 1)}{(\sqrt{5} + 1)} = \frac{(\sqrt{5} + 1)}{5-1} = \frac{(\sqrt{5} + 1)}{4} $

The original form required dividing by $(2.23606798... -1)$, not easy, while the new format (again, looks different but is in fact equal) is divided by 4, much easier and simpler.

Now that we have clarified what we are doing and why, your problem is simple. Here we are told to rationalize the numerator -- less common than the denominator but useful at times:

$\frac{\sqrt{a} - 2}{a - 4} = \frac{\sqrt{a} - 2}{a - 4} * 1 = \frac{\sqrt{a} - 2}{a - 4} * \frac{\sqrt{a} + 2}{\sqrt{a} + 2} = \frac{(a-4)}{(a-4)(\sqrt{a} + 2)} = \frac{1}{(\sqrt{a} + 2)}$

Note of course a cannot be equal to 4 so this fraction can be defined.

Practice this technique. It is quick and easy and has many applications.

Answer 2

So $a - 4 = (\sqrt a - 2)(\sqrt a + 2)$, giving you a fraction of

$$ \frac{\sqrt a - 2}{ (\sqrt a - 2)(\sqrt a + 2)} = \frac{1}{\sqrt a + 2}$$ as desired.

This turns out to be a problem where you recognize that the denominator can be factored by something in the numerator - basically it's a pattern recognition problem.

Answer 3

Rationalizing the numerator works the same way as rationalizing the denominator, i.e.

$$\ \dfrac{\sqrt a -2}{b}\ =\ \dfrac{\sqrt a -2}{b}\, \dfrac{\sqrt a +2}{\sqrt a + 2}\ =\ \dfrac{a-4}{b(\sqrt a + 2)}$$

OP is the special case $\, b = a\!-\!4,\,$ so they cancel out $ $ (note $\,\sqrt a + 2\neq 0\,$ by $\,\sqrt a \ge 0)$

Remark $\ $ This often proves handy, e.g. we can make the quadratic formula work even in the degenerate case when the leading coefficient $\,a = 0\,$ by rationalizing the numerator as below

$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ =\ \dfrac{2c}{-b \pm \sqrt{b^2-4\:a\:c}}$$

As $\,a\to 0,\,$ the latter yields the root $\,x = -b/c\ $ of $\ bx+c\,\ (= ax^2\!+bx+c\,$ when $\,a=0).$

Similarly it often proves useful for evaluating various limits.