How to simplify radicals

[GrannySmith]

How to simplify radicals in denominator

I'm in algebra 2. Looking for some math help and found these forums . Sorry if I type some of the problems wrong I have no idea how to type some of this math stuff on computer.

I need help with simplifying radicals in denominator. I know how to do some of these problems, but I need help especially when the denominator has two terms or imaginary numbers in the radicals.

Here are some of the problems I'm stuck on. As far as I'm concerned the object is to get rid of all radicals in the denominator correct? I understand how to do some of these problems and would like to see if my answers are correct. On other problems, I honestly have no clue how to start!

1. √9 / 6√8 Answer i got: 1 / 4√2

2. √3 / 7√5 Answer I got √15 / 35

3. 8√6 + √10 / 2√2 - √7

4. 5 + i / 4 + 3i

5. 7√72 / √40 Answer I got: 21√5 / 5

1 and 2 seem pretty easy to me. I just multiply the radical to both sides to get rid of it. For 4 I have no idea how to do it when there are imaginary numbers lol.

Thanks in advance for the help!

 

[Bob Brown MSEE]

1. √9 / 6√8 Answer i got: 1 / 4√2

Correction: Format
1. √9 / 6√8 Answer i got: 1 / 4√2 Wrong
1. √9 / 6/√8 Answer i got: 1 / 4/√2 Correct
1. √9 / (6√8) Answer i got: 1 / (4√2) Correct

Correction: Answer
After the Format is corrected, the answer is not in the form required by the problem statement,
you still have a radical in the denominator.

 

[GrannySmith]

Correction: Format
1. √9 / 6√8 Answer i got: 1 / 4√2 Wrong
1. √9 / 6/√8 Answer i got: 1 / 4/√2 Correct
1. √9 / (6√8) Answer i got: 1 / (4√2) Correct

Correction: Answer
After the Format is corrected, the answer is not in the form required by the problem statement,
you still have a radical in the denominator.

I understand what you're saying, but what difference does the parenthesis make? Isn't that still a radical?

I redid the problem and got 1/8. What I did was simplified 4√2 by multiplying both sides by √2. That makes the top √2 and the bottom 2 times 4 which is 8.

I got 4√2 by simplifying 6√8

 

[Quaid]

Your answers for #2 and #5 are correct.

Have you learned about conjugates?

Multiply top and bottom of #3 by 2√2 + √7

Multiply top and bottom of #4 by 4 - 3i



PS: Don't forget grouping symbols, when texting algebraic fractions. For example,

√9 / 6√8 should be typed as √9 / (6√8)

8√6 + √10 / 2√2 - √7 should be typed as (8√6 + √10) / (2√2 - √7)

Let us know, if you're not sure why. Cheers :cool:

 

[GrannySmith]

Now that I sort of understand the general concept behind solving these equations, where should I post if I have any specific questions?

Like sometimes, I'll get those problems where I just don't know what to do for some reason, even though the answer may be really easy. They just stump me and I don't know how to proceed, but when it is explained to me I completely understand the whole thing.

Thank you so much for all the help! Finally starting to understand this now, but I've run into another bump.

1. (10 + √10)/(-5 + √5) By multiplying both sides by the conjugate, I get a very large fraction. (-50 - 10√5 - 5√10 - √50)/30 Is this the simplified answer? I do not know how to proceed. They don't have a common factor. because of the √50 that is alone.

2. Dealing with imaginary numbers. (5 + 5i)/(1 + 10i) I got (-45 + 55i)/ -99 and it just doesn't feel right. Nothing can be simplified further correct?

 

[Quaid]

where should I post if I have any specific questions?

You're already posting in a good place. Please start a new thread for each new topic or exercise; threads become less cluttered, that way.

sometimes ... They just stump me and I don't know how to proceed, but when it is explained to me I completely understand the whole thing.

That's called learning. The more you practice, the more you see. The more you see, the more easily you move forward...

(10 + √10)/(-5 + √5) By multiplying both sides by the conjugate, I get a very large fraction. (-50 - 10√5 - 5√10 - √50)/30

They don't have a common factor. because of the √50

You did pretty good, but the denominator is not correct. (Typo, maybe?) Please check your calculation; the correct value is a multiple of 10, but it's not 30.

Also, if you reduce √50, then each term in the numerator will have a common factor, and that allows you to simplify somewhat.

Show us what you get, even if you get stuck. We will go from there.

(5 + 5i)/(1 + 10i) I got (-45 + 55i)/ (-99) and it just doesn't feel right.

Well, your gut instinct is working.

Again, the denominator is not correct. Fix that, and I think you're done.

(Also, it's good form to enclose negative denominators inside grouping symbols, as shown in red.)


Cheers :cool:

 

[GrannySmith]

If my denominator is the only thing that is incorrect, perhaps I am distributing the conjugate wrong?

My teacher taught me that the shortcut for the bottom term would just be to square it since you're basically multiplying it by itself anyways. When I square -5 + √5 I get 25 + 5 which is 30. I'm not sure if this is my mistake, but perhaps I forgot to change the sign and the real answer is 20?

As for the top when I simplify the √50 I get (-10 - 2√5 - √10 - √2). If in fact the denominator is 20 I can reduce each term by 5 which gets me (-10 - 2√5 - √10 - √2)/4. Is this the final correct answer?

For my second problem, I'm going to assume I made the same mistake when "squaring" the bottom term. Instead of getting 1 - 100 which gets me (-99) which is wrong according to you, I should be getting 1 + 100 = 101?

Final answer for that would be (-45 + 55i/101)?

I hope I have found the stupid mistake that has been messing up all of my problems lately. If I happen to come across any other bumps I should make a new thread instead of posting in this one correct? Lastly, thank you so much for your help! Wouldn't have gotten through this without you guys. It all seems much simpler now.

 

[Quaid]

My teacher taught me that the shortcut for the bottom term would just be to square it since you're basically multiplying it by itself anyways.

That's not the case when the denominator is a sum or difference of terms containing either a radical or an imaginary number.

In these cases, you need to multiply both the numerator and denominator by the conjugate of the denominator.

(-5 - √5) (-5 + √5) = 20

(1 + 10i) (1 - 10i) = 101

As you see, this is not squaring. Use the FOIL algorithm, to multiply these. (You've learned FOIL, yes?)

As for the top when I simplify the √50 I get [this factorization for the numerator:]

5 (-10 - 2√5 - √10 - √2)

If in fact the denominator is 20 I can [cancel the factor of 5 in the numerator with a factor of 5 in the denominator]

which gets me (-10 - 2√5 - √10 - √2)/4. Is this the final correct answer?

Yes, that's good.

Some teachers may prefer not seeing a negative first-term in a numerator, so we might also factor out -1 from each term on top.

-(10 + 2√5 + √10 + √2)/4


Note: Edits shown in blue are what I think you were trying to say.

For my second problem, I'm going to assume I made the same mistake

I should be getting 1 + 100 = 101?

Final answer for that would be (-45 + 55i/101)?

Yes and yes.

Also, the grouping symbols should enclose only the numerator, not the entire ratio.

(-45 + 55i)/101

Again, if teacher prefers:

-(45 - 55i)/101

or

(55i - 45)/101


All different forms of the same ratio...

If I happen to come across any other bumps I should make a new thread instead of posting in this one correct?

If the post concerns a different topic or a new exercise, then please start a new thread. If questions remain about the exercises in this thread, then post here.

Cheers :cool:

 

[GrannySmith]

Yes one last type of problem.

If the denominator has only one term, a monomial how would I solve it?

For example 4/(-6i). I multiply both sides by 6i to get rid of the i on the bottom.
This gives me (-24i)/36 which I can simplify down to -4i/6 correct?

Another problem is 2/7i. This one I did the same thing and got 2i/-49 but my friend is telling me he got -2i/7. No idea how he got that and I don't understand. Is he correct or am I correct?

 

[JeffM]

Yes one last type of problem.

If the denominator has only one term, a monomial how would I solve it?

For example 4/(-6i). I multiply both sides by 6i to get rid of the i on the bottom.
This gives me (-24i)/36 which I can simplify down to -4i/6 correct? No.

Another problem is 2/7i. This one I did the same thing and got 2i/-49 but my friend is telling me he got -2i/7. No idea how he got that and I don't understand. Is he correct or am I correct? He is.

Rationalization involves multiplying by 1.

\(\displaystyle \dfrac{4}{-6i} = \dfrac{4}{-6i} * 1 = \dfrac{4}{-6i} * \dfrac{i}{i} = \dfrac{4i}{-6i^2} = \dfrac{4i}{-6(- 1)} = \dfrac{4i}{6} = \dfrac{2i}{3}.\)

\(\displaystyle \dfrac{2}{7i} = \dfrac{2}{7i} * 1 = \dfrac{2}{7i} * \dfrac{i}{i} = \dfrac{2i}{7i^2} = \dfrac{2i}{7(-1)} = \dfrac{2i}{- 7} = \dfrac{-2i}{7}.\)