How do you rationalize a monomial denominator?

[lj19]

How do you rationalize a monomial denominator? [Algebra 2/Trig.]

Could someone teach me how to solve these types of equations?

For example, radical 5 over radical 3.

 

[fzero]

Is $$\sqrt{3}\sqrt{3}$$ rational?

 

[lj19]

No I don't think so.

 

[cronxeh]

Well.. sqrt(5)/sqrt(3) you can multiply it by sqrt(3) on top and bottom, sqrt(3)*sqrt(5)/sqrt(3)^2 = sqrt(3)*sqrt(5)/3 so now you can go in for combination. so now you have sqrt(15)/3

So easiest way to do this.. is divide 15 by 9 and take a square root of that. So now you have sqrt(1.666...) (which if you notice you could've gotten by taking sqrt(5/3)

Do you know how to take a radical by hand?

 

[lj19]

We did this problem in class, but I have some questions.
How did you know to multiply by sqrt[3] over sqrt[3], is it because then you have a common denominator?
Could you please explain this part "sqrt(3)*sqrt(5)/sqrt(3)^2 = sqrt(3)*sqrt(5)/3"?

 

[cronxeh]

We did this problem in class, but I have some questions.
How did you know to multiply by sqrt[3] over sqrt[3], is it because then you have a common denominator?
Could you please explain this part "sqrt(3)*sqrt(5)/sqrt(3)^2 = sqrt(3)*sqrt(5)/3"?

Well generally you don't want to leave a radical in the denominator, so you should just leave a number there and radicals in the numerator. Sometimes you get lucky and the number is simple but often times its not.

sqrt(3)*sqrt(5) is your numerator after you've multiplied it out, and on the denominator you have sqrt(3)*sqrt(3) which is sqrt(3)^2 or $$\sqrt{3}$$² which is just 3

 

[lj19]

I understand that you cannot have a radical in the denominator, because then it isn't simplified. Why is the sqrt[3] just 3, isn't that still a radical?

 

[cronxeh]

I understand that you cannot have a radical in the denominator, because then it isn't simplified. Why is the sqrt[3] just 3, isn't that still a radical?

You multiplied both top and bottom by sqrt(3). if you multiplied sqrt(3) by sqrt(3) you get 3

$$\frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{3} \sqrt{5}}{\sqrt{3} \sqrt{3}} = \frac{\sqrt{15}}{3}$$

 

[lj19]

I'll solve the problem the way you explained and see if it makes sense.
My teacher explained it by taking the original problem of radical 5/radical 3 and multiplying it by radical 3/radical 3. Then she wrote that from that you get just radical 9 in the denominator, and then radical 15 over 3. Is that correct? Why between the final answer of radical 15/3 and the radical 3/radical 3 did she just get a radical 9 in the denominator?
Thanks.

 

[lj19]

I think I understand why. In another problem that is sqrt 3 over sqrt 7 she multiplied it by sqrt 7 over sqrt 7 [because then it's a common denominator...?] And then put just 49 in the denominator as the next step. And the final answer was sqrt 21/7. So that step of radical 49 [or "radical 9" above] is just a simplifying step?

 

[cronxeh]

I'll solve the problem the way you explained and see if it makes sense.
My teacher explained it by taking the original problem of radical 5/radical 3 and multiplying it by radical 3/radical 3. Then she wrote that from that you get just radical 9 in the denominator, and then radical 15 over 3. Is that correct? Why between the final answer of radical 15/3 and the radical 3/radical 3 did she just get a radical 9 in the denominator?
Thanks.

Because she is boring.

The real math here is just taking $$\sqrt{5/3}$$ and not thinking about it twice.

 

[cronxeh]

I think I understand why. In another problem that is sqrt 3 over sqrt 7 she multiplied it by sqrt 7 over sqrt 7 [because then it's a common denominator...?] And then put just 49 in the denominator as the next step. And the final answer was sqrt 21/7. So that step of radical 49 [or "radical 9" above] is just a simplifying step?

Its just a confirmation step that you know what you doing. sqrt(7)*sqrt(7) = sqrt(7*7) = sqrt(49) = 7

Do you want to learn how to take a square root by hand?

 

[lj19]

Okay I understand that is is a simplifying step, only using the denominator.
I had one more questions about this problem, which is with any number in any monomial denominator problem you have to rationalize is the first step to take the original problem and multiply it by it's denominator over that same number?

 

[cronxeh]

Okay I understand that is is a simplifying step, only using the denominator.
I had one more questions about this problem, which is with any number in any monomial denominator problem you have to rationalize is the first step to take the original problem and multiply it by it's denominator over that same number?

No like I said its just a teaching methodology. The trick here is that you can combine radicals in division and multiplication. sqrt(a)*sqrt(b) = sqrt(a*b). sqrt(a)/sqrt(b) = sqrt(a/b) provided a and b are both positive (a,b > 0)

 

[lj19]

So then you would always use that denominator. For example, if the problem was sqrt[7]/sqrt[4] you would multiply it by sqrt[4]/sqrt[4]?

And then after multiply that, and simplify it [and have no radical in the denominator].

 

[cronxeh]

So then you would always use that denominator. For example, if the problem was sqrt[7]/sqrt[4] you would multiply it by sqrt[4]/sqrt[4]?

And then after multiply that, and simplify it [and have no radical in the denominator].

It depends on what you want. If you just want to simplify the problem, then don't leave a radical in the denominator, so multiply the denominator by numerator and denominator.

If you want to get the answer as in by hand, then just combine the terms and long divide sqrt(4/7)

 

[lj19]

I understand. I can solve it how I described, by multiplying, and getting radical 16 in the denominator and then getting sqrt28/4.

Could someone also explain how to rationalize the binomial denominator of a radical expression?

Thank you.

 

[fzero]

I understand. I can solve it how I described, by multiplying, and getting radical 16 in the denominator and then getting sqrt28/4.

Could someone also explain how to rationalize the binomial denominator of a radical expression?

Thank you.

You just use the same idea. If by binomial you mean something like

$$\sqrt{ \frac{x+3}{x+7} } = \frac{\sqrt{x+3}}{\sqrt{x+7}} \frac{\sqrt{x+7}}{\sqrt{x+7}} = \frac{\sqrt{(x+3)(x+7)}}{x+7}.$$

 

[lj19]

We learned how to in class so I have notes on it, although it is very complicated. I notice that for example one problem is 9/(5-sqrt13) and then you multiply it by the denominator [to get a common denominator] of (5 PLUS [sign changes] sqrt13)/(5+sqrt13). Then you multiply those, and get 25+5sqrt13-5sqrt73-13 [that is the denominator part simplified]. And then it is simplified again and then the final answer is solved. But how does the denominator, 5 minus radical 13 multiplied by 5 plus radical 13 equal 25 plus 5 times radical 13 minus -5 times radical 73 minus 13?
If you multiply the denominators across as I just explained, it does not equal that??"And then it is simplified again and then the final answer is solved."
After that the denominator is simplified further to 25-13 and then the answer is 9(5+8sqrt73)/12, which becomes 3(5+8sqrt73)/4.

 

[fzero]

Well, it's simple to verify that (and well worth remembering, since it comes up a lot)

$$(x+y)(x-y) = x^2-y^2.$$

Therefore, if we have an expression $$a+\sqrt{b}$$, we can multiply it by $$a-\sqrt{b}$$ to obtain

$$(a+\sqrt{b})(a-\sqrt{b})=a^2 - b.$$

So if we had an expression with $$a+\sqrt{b}$$ in the denominator, we can multiply by

$$\frac{ a-\sqrt{b}}{a-\sqrt{b}}.$$

to rationalize the denominator.

 

[lj19]

In this problem the denominators being multiplied are (5-sqrt13)(5+sqrt13). That answer is to 25+5sqrt13-5sqrt73-13.
But when you multiply those do you use the FOIL method? Or multiply reguarily?

 

[fzero]

In this problem the denominators being multiplied are (5-sqrt13)(5+sqrt13). That answer is to 25+5sqrt13-5sqrt73-13.
But when you multiply those do you use the FOIL method? Or multiply reguarily?

FOIL is just a mnemonic for what I'd call multiplying correctly, but so as not to be confused, you can always use FOIL here. But the outside and inside terms cancel when you multiply (x+y)(x-y) = x² - xy + yx - y².

Also I messed up some signs in my previous post that I'll go back and fix.

 

[lj19]

It wasn't in my notes that I had to use the FOIL method, but now the problem makes sense.
After you multiply the original problem [9/5-sqrt13] by the denominator/demonimator [5+sqrt13/5+sqrt13] and then multiply the denominators (5-sqrt13)(5+sqrt13) to get 25+5sqrt13-5sqrt73-13 then the denominators are simplified to 25 minus 13. How is 25+5sqrt13-5sqrt13-13 simplifed to 25 minus 13?
Thank you.

 

[fzero]

How is 25+5sqrt13-5sqrt73-13 simplifed to 25 minus 13?
Thank you.

You have a typo, there should be a 13 instead of 73 in the 3rd term.

 

[lj19]

Thank you!
I need to take neater notes!
Then it becomes 12 on the denominator after you subtract 13 from 25. Then the problem is 9(5+sqrt13)/12 which simplifies to 3(5sqrt13)/4 as the answer.

Originally the numerator is 9 and then is multiplied by the 5+sqrt13 which equals 9(5+sqrt13) before simplified to the final answer.
Is it because an addition symbol is present in the problem that you cannot combine it?

 

[fzero]

Thank you!
I need to take neater notes!
Then it becomes 12 on the denominator after you subtract 13 from 25. Then the problem is 9(5+sqrt13)/12 which simplifies to 3(5sqrt13)/4 as the answer.

I have one more question about this problem. I understand it now. But if originally the numerator is 9 and then is multiplied by the 5+sqrt13 why does that equal 9(5+sqrt13) before simplified to the final answer.
Why is 9 multiplied by 5+sqrt13 just 9(5+sqrt13)? Is it because the 9 cannot be combined with the 5+sqrt13 because of the addition symbol?

Multiplication is distributive with respect to addition, which means that c(a+b) = c a + c b. In your case

$$9(5+\sqrt{13}) = (9)(5)+ 9\sqrt{13} = 45+9\sqrt{13}.$$

 

[lj19]

Nevermind I understand how I did the problem.

 

[lj19]

Can you briefly explain how to divide monomials.
From my notes a problem is, (sqrt16)/(sqrt2) which simplifies to sqrt16/2 which simplifies to sqrt8 which simplifies to sqrt4(2) and then answer is 2(sqrt2).
Can you explain this problem and how to divide monomials?

In another problem from my notes on dividing monomials is, (10(sqrt36))/(2(sqrt3)) which simplifies to (5(sqrt12))/(2(sqrt3)) which simplifies to 10sqrt3. The sqrt12 has a 2 on the index, my teacher said that if the number of the index is not given, then it is always 2. Could someone explain that to me?

 

[fzero]

Can you briefly explain how to divide monomials.
From my notes a problem is, (sqrt16)/(sqrt2) which simplifies to sqrt16/2 which simplifies to sqrt8 which simplifies to sqrt4(2) and then answer is 2(sqrt2).
Can you explain this problem and how to divide monomials?

The square root distributes over products and quotients, so in general

$$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}},$$

Therefore we apply this in reverse to

$$\frac{\sqrt{16}}{\sqrt{2}} = \sqrt{\frac{16}{2}}.$$

In another problem from my notes on dividing monomials is, (10(sqrt36))/(2(sqrt3)) which simplifies to (5(sqrt12))/(2(sqrt3)) which simplifies to 10sqrt3. The sqrt12 has a 2 on the outside of the radical symbol, my teacher said that if the number on the outside of the radical symbol isn't given, then it is always 2. Could someone explain that to me?

The square root symbol is used to represent the inverse of the square function, that is if

$$y = x^2,$$

then

$$x = \sqrt{y}.$$

We can also raise x to some other power n, in that case we use the radical with the number on the outside to represent the inverse:

$$y = x^n, ~~ x = \sqrt[n]{y}.$$

To be specific, if x = 2 and n=3, we have

$$8 = 2^3, 2 = \sqrt[3]{8}.$$

 

[lj19]

I understand that (sqrt16)/(sqrt2) simplifies to sqrt16/2 and simplifies to sqrt 8.
Why does sqrt8 simplify to sqrt4(2)? Which then simplifies to 2(sqrt2).

 

[eumyang]

Why does sqrt8 simplify to sqrt4(2)? Which then simplifies to 2(sqrt2).

We use the product property of radicals:
$$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$$
... so
$$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}$$

 

[lj19]

I have a question about radicals in general.
If you are adding [or subtracting] radicals or radicals that are being multiplied by a number, the radical stays the same correct? For example sqrt5+sqrt5=sqrt5? Another example is 4(sqrt5)+2(sqrt5), where the sqrt5 would stay sqrt5.
If you are multiplying [or dividing] radicals or radicals that are being multiplied by a number, the number in the radical is multiplied correct? For example, (5(sqrt10))(1(sqrt10))=5(sqrt100).
Is this correct?

 

[fzero]

I have a question about radicals in general.
If you are adding [or subtracting] radicals or radicals that are being multiplied by a number, the radical stays the same correct? For example sqrt5+sqrt5=sqrt5? Another example is 4(sqrt5)+2(sqrt5), where the sqrt5 would stay sqrt5.
If you are multiplying [or dividing] radicals or radicals that are being multiplied by a number, the number in the radical is multiplied correct? For example, (5(sqrt10))(1(sqrt10))=5(sqrt100).
Is this correct?

$$\sqrt{5}+\sqrt{5} = 2\sqrt{5}.$$

$$4\sqrt{5}+2\sqrt{5} = 6\sqrt{5}.$$

These are both examples of the distributive law for multiplication ac + bc = (a+b)c.

$$(5\sqrt{10})(1\sqrt{10}) = 5 \sqrt{100}$$

uses the associative law a(bc) = (ab)c and the commutative law ab=ba, as well as the product property of radicals.

 

[lj19]

You're correct, I forgot that there is always a 1 in front of the radical. So it's correct that when adding radicals the number in the radical does not change, but when multiplying the number in the radical is multiplied?

Could you help me understand how to solve problems with the index with radicals.
For example, [multiplying a monomial] 4(sqrt of x to the 5th power) with an index of 3 in the radical, multiplied by sqrt of 16 x to the second power with an index of 3 in the radical.
Also, how do I write this mathematically on this website?
In my notes this problem was simplified to 4(16x to the seventh power) with a 3as the index, and then simplified to 4(8x to the sixth power times 2x) with a 3 as the index.
How would I solve this problem?

 

[fzero]

You're correct, I forgot that there is always a 1 in front of the radical. So it's correct that when adding radicals the number in the radical does not change, but when multiplying the number in the radical is multiplied?

Instead of asking when the number changes, you should review the distributive law that I've been quoting and the product rule for radicals. What you say is roughly true but you can make other choices for how to express a sum or product. For example,

$$\sqrt{5} + \sqrt{5} = 2 \sqrt{5} = \sqrt{20}$$

$$(5\sqrt{10})(\sqrt{10}) = 5 \sqrt{100} = \sqrt{2500} = 50.$$

Rather than sticking to rules about whether the radical changes or not, you should learn to become comfortable with multiplication and exponents.

Could you help me understand how to solve problems with the index with radicals.
For example, [multiplying a monomial] 4(sqrt of x to the 5th power) with an index of 3 in the radical, multiplied by sqrt of 16 x to the second power with an index of 3 in the radical.

In my notes this problem was simplified to 4(16x to the seventh power) with a 3as the index, and then simplified to 4(8x to the sixth power times 2x) with a 3 as the index.
How would I solve this problem?

Do you mean

$$(4 \sqrt[3]{x^5})( \sqrt[3]{(16 x)^7}) ?$$

I would put everything under the radical and then express everything in terms of powers of 2 and x:

$$(4 \sqrt[3]{x^5})( \sqrt[3]{(16 x)^7}) = \sqrt[3]{(4^3)x^5( (2^4) x)^7} =\sqrt[3]{(2^6)(2^{28}) x^{12}} = \sqrt[3]{(2^{34})x^{12}} = 2^{11} x^4\sqrt[3]{2} .$$

I don't know if that's the same result that you got in class, but it's the simplest form for the expression, assuming that I didn't make any mistakes.

Also, how do I write this mathematically on this website?

There's some references on how to write the LaTeX code in this post:

https://www.physicsforums.com/showthread.php?t=386951

It takes some getting used to.