# توجیه کردن

سرفصل: بخش ریاضی / سرفصل: ریشه ها و قدرت ها / درس 15

## بخش ریاضی

14 سرفصل | 192 درس

### توضیح مختصر

• زمان مطالعه 13 دقیقه
• سطح خیلی سخت

### دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید ## Rationalizing

Now we can talk about the very important topic of rationalizing. So throughout this module, we have seen a few fractions involving radicals. Notice that so far, no final answer has involved a radical in the denominator. And in fact I have crafted every fraction to make sure that this never happened, that it never was a result that a radical was left over in the denominator. Now why would I do that?

This is a tricky issue. Having a radical in the denominator is considered in poor taste, in mathematics. Now, why would mathematicians think this? One way to think about it is, if there are multiple fractions that need to be added or subtracted it’s helpful if all the denominators are integers so that we can find a common denominator.

A more general way to say it is. If there are multiple ways to write an answer, it’s helpful if there’s an established convention so that everyone writes the answer the same way, and that makes it much easier to compare whether two different people have found the same answer. So for whatever reason, the mathematical convention is to avoid radicals in the denominator.

Now, if a radical appears in the denominator as part of the problem, and of course if sometimes happens, we need to change the fraction to an equivalent form, so that no radicals appears in the denominator. This process of change is called rationalizing the denominator. That is to say, making the denominator a rational func, a rational number, an integer rather than an irrational square root.

So, suppose in the course of solving a problem, we found x equals 12 divided by root three. Let’s pretend that we did all the math correctly, we got this as our correct answer. This is our solution to the problem. That might be mathematically correct, but because it’s not rationalized, we won’t find it among the answers listed in this, won’t find the answer listed in that particular form.

You see all the answers on the test are rationalized, so we need to rationalize anything that we find to match it to the answer choice. If there’s a single radical in the denominator, we rationalize simply my multiplying by that radical over itself. So, we’re just gonna multiply by the square root of three, over the square root of three.

And of course, in the denominator, we get root three, twelve root three. Well, think about this. Root three is the number which, when we square it, we get three. So, if here, we multiply it by itself, so it becomes three. Then we can cancel the 12 divided by three. That becomes four.

We get four root three. That is the rationalized form, so the answer would never be written in the form 12 divided by root three. It would be written in the form four times root three. Practice rationalizing the following.

In each case you’re just gonna multiply by the radical over itself and then simplify. Pause the video and then we’ll talk about this. Okay and that first one obviously we will multiply by root five over root five and we get root five over five. In the second one, we’ll multiply by root three over root three. We’ll wind up with two root three over three.

In the third one, we’ll multiply by root 21 over root 21. We get a 21 in the denominator, that allows to cancel a factor of seven. And wind up with two root 21 over three. Those are the rationalized forms. Even if there is addition or subtraction in the numerator divided by a single root, all we have to do is multiply by the root over itself.

So, for example if we had something like four minus root six, over two root three. Really the only thing that concerns us is that root three in the denominator. So I’m gonna multiply by root three over root three. In the denominator I get a root three times a root three, which is three, and the number I distribute. That root six times root three.

Remember that root six is root two times root three. So we can multiply the two factors of root three together to get three. And then we’re just left with a root two left over. So what we wind up with, is four root three, minus three root two, over six. And that is the fully rationalized form. Things get trickier when we have addition or subtraction in the denominator.

So suppose we have to simplify something like two divided by the quantity, root five minus one. Well, hm. What we have been doing is not gonna work here. If see, if we multiply this denominator by root, by this root five, we would have to distribute across the to subtract, across the subtraction in the denominator.

The root five times five, that would also be five. But, of course, they’ll also be a one times root five. And that would be root five. And our attempt to rid the denominator of radicals would not be successful. We have to make use of a different trick, so I will say remember the difference of two squares formula from the algebra module.

P minus q times p plus q equals p squared minus q squared. If this is an unfamiliar formula I highly suggest go back to the algebra modules and watch the videos concerning this particular formula. It’s one of the most widely applicable formulas in all of mathematics. So in other words what we have here is the product of a sum and the, and the difference, as a result of squaring both terms.

If p or q, or even both of them, were radical expressions, then on the right side of the equation above, every radical expression would be squared, and there would be no radicals in the result. Very interesting. Consider any two terms, any two term expression, either addition or subtraction, in which one or more of the terms is a radical.

So we might have things like this for example. Two terms, and we’re either adding or subtracting, and one or both of the terms is a radical. If we change the addition or subtraction to its opposite, we construct what is called the conjugate expression. So that first one we have three minus root seven, the conjugate would be three plus root seven.

In the second one, we have root 13 plus two root 11. The conjugate will be root 13 minus two root 11. And so how is this helpful? Well, if we multiply any expression with radicals by its conjugate, one conjugate plays the role of a plus b, the other plays the role of a minus b. So that their product is the difference of the squares of the two terms, a squared minus b squared.

So for example if we had three root seven, three minus root seven times three plus root seven. Well essentially what we have here is an a minus b times and a plus b and of course this is gonna equal a squared minus b squared. Or in other words three squared minus root seven squared. Square each on of them we get 9 minus 7 which is 2.

So, in other words, we multiply two expressions with radicals and we get a product without radicals. That’s a really big idea. This provides a clue we need for how to rationalize a fraction with a radical expression involving addition or subtraction in the denominator. So let’s go back to that one that we couldn’t solve a moment ago.

We had something like two divided by the quantity root five minus one. Well what I’m gonna do is multiply the numerator and denominator by the conjugate of the denominator. To rationalize, we’re gonna multiply the numerator and the denominator by the conjugate of the denominator. So what we have in that denominator is root five minus one, the conjugate would be root five plus one.

So we’re gonna multiply by that conjugate, root five plus one over root five plus one. Then in that denominator, we have an a minus b and an a plus b so we’ll get an a squared minus b squared in that denominator. The numerator I’m just going to leave undistributed for a moment. And so we get a root five squared minus 14 squared.

That would be five minus one, which is four. Then I can cancel a factor of two in the numerator and denominator. And I get root five plus one over two, and that is the rationalized form. Incidentally, you don’t have to know this for the test, but that happens to be the golden ratio. So rationalize the following.

You will need to multiply by the denom, multiply the denominator by its conjugate. So I’m gonna say pause the video and try this on your own. Okay, so we have a root seven minus root three, and so what we’re gonna need to do is multiply that denominator by our root seven plus root three. So we have to multiply root seven plus root three over root seven plus root three, multiplied by the same thing over itself.

In the numerator, I’m not going to distribute just yet. In the denominator, I get the square of those terms, the difference of the squares, which just seven minus three. Seven minus three is four. And then cancel eight divided by four and we get two times root seven plus root three, and that is the fully rationalized form.

Now we can distribute it if we want to depending, doesn’t matter, the answer could be written in either of those last two forms. Rationalize the following. You will need to multiply the denominator by its conjugate. Pause the video work on this and then we’ll talk about it. Okay so we have a three plus root five in the denominator so we need to multiply by a three minus root five.

So we need to multiply both the numerator and the denominator by three minus root five. In the denominator we’re just gonna get the difference of two squares. Three squared minus root five squared. In the numerator I’m gonna have to foil. And so what I get is the product of the first, four times three is 12.

The product of the outer, four so that’s minus four, root five. The product of the inner. That’s plus six root five. And then the product of the last, which is two root five minus negative root five. Well the root five times the root five is five, two times five is ten so that’s just minus ten.

Now we’ll simplify everything in the numerator and the denominator. We wind up with two plus two with five over four. Everything is divisible by two, so we can divide it by a factor of two. And we are left with one plus root five divided by two. Here’s a practice problem.

Pause the video, and then we’ll talk about this. Okay, so we need to get x by itself, so we are, of course we are going to divide by everything in the parenthesis. We have an expression in the denominator, involving addition or subtraction with a radical.

So we’re gonna multiply by the conjugate. We have a two root five minus three. We need to multiply that by a two root five plus three. And so I’ll multiply by that. In the numerator I’ll just leave it undistributed. In the denominator, I’m gonna get two root five.

That thing squared minus three root squared. And, of course, two root pi squared as we learned from our operations of radicals a few videos ago. That’s two squared tim, times root five squared. Four times root five, four times five which is 20.

And of course three squared is nine. 20 minus nine we get 11. We can cancel a factor of 11 in the numerator, denominator and we’re left with five times the quantity two root five plus three and that is the fully rationalized form of the answer. In summary to eliminate roots from the denominator of a fraction this is called rationalizing and it’s something you always have to do.

The test answers, the answers that appear on multiple choice will always be rationalized and so we will have to rationalize to match our answers to this. If the fraction has a single root in the denominator, we rationalize simply by multiplying by that root over itself. If the denominator of the fraction contains addition or subtraction involving a radical expressions, to rationalize we need to multiply by the conjugate of the denominator over itself.

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