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How to simplify radicals in the denominator?

GrannySmith

New member
Jan 5, 2014
5
I have some problems I am stuck on. The goal here is to simplify radicals in the denominator. I understand that when there is a binomial in the denominator, you need to multiply both sides by the conjugate. For some reason though, I seem to be having trouble doing that or am making a mistake somewhere that I cannot figure out.

1. (8√6 + √10)/(2√2 - √7)

I multiply both sides by the conjugate and get (16√12 + 8√42 + 2√20 +√70)/1. Then when I simplify this I get 33√3 + 8√42 + 4√5 + √70. Seems like a long answer. What am I doing wrong?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are not expanding correctly. The second binomial has a negative sign that you aren't taking into account. Recall:

\(\displaystyle (a+b)(c-d)=ac-ad+bc-bd\)
 

GrannySmith

New member
Jan 5, 2014
5
I redid the problem like 2 times again but cannot catch on to my mistake. My mistake is in the denominator correct?

(2√2 - √7)(2√2 +√7)

I'm going to take this step by step.

(2√2)(2√2) = 4√4 = 8

(2√2)(√7) = 2√14

(-√7)(2√2) = -2√14

(-√7)(√7) = -√49 = -7

2√14 - 2√14 cancel each other out.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I have some problems I am stuck on. The goal here is to simplify radicals in the denominator. I understand that when there is a binomial in the denominator, you need to multiply both sides by the conjugate. For some reason though, I seem to be having trouble doing that or am making a mistake somewhere that I cannot figure out.

1. (8√6 + √10)/(2√2 - √7)

I multiply both sides by the conjugate and get (16√12 + 8√42 + 2√20 +√70)/1. Then when I simplify this I get 33√3 + 8√42 + 4√5 + √70. Seems like a long answer. What am I doing wrong?
That looks correct to me.
 

GrannySmith

New member
Jan 5, 2014
5
That looks correct to me.
I redid this problem and cannot find anything wrong.

It's just that i have this gut feeling that this answer is wrong for some reason. Way longer than any of my past answers for other problems like this. If it's right though then I guess there's nothing I can do to simplify that!
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
I have some problems I am stuck on. The goal here is to simplify radicals in the denominator. I understand that when there is a binomial in the denominator, you need to multiply both sides by the conjugate. For some reason though, I seem to be having trouble doing that or am making a mistake somewhere that I cannot figure out.

1. (8√6 + √10)/(2√2 - √7)

I multiply both sides by the conjugate and get (16√12 + 8√42 + 2√20 +√70)/1. Then when I simplify this I get 33√3 + 8√42 + 4√5 + √70. Seems like a long answer. What am I doing wrong?
Why do you think having a long answer makes it incorrect?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have some problems I am stuck on. The goal here is to simplify radicals in the denominator. I understand that when there is a binomial in the denominator, you need to multiply both sides by the conjugate. For some reason though, I seem to be having trouble doing that or am making a mistake somewhere that I cannot figure out.

1. (8√6 + √10)/(2√2 - √7)

I multiply both sides by the conjugate and get (16√12 + 8√42 + 2√20 +√70)/1. Then when I simplify this I get 33√3 + 8√42 + 4√5 + √70. Seems like a long answer. What am I doing wrong?
You are not expanding correctly. The second binomial has a negative sign that you aren't taking into account. Recall:

\(\displaystyle (a+b)(c-d)=ac-ad+bc-bd\)
My apologies...I misread the expression you gave as the product of numerator and the conjugate, not the original. :eek:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I redid this problem and cannot find anything wrong.

It's just that i have this gut feeling that this answer is wrong for some reason. Way longer than any of my past answers for other problems like this. If it's right though then I guess there's nothing I can do to simplify that!
The radicals in the original problem all contain different prime factors (3, 5, 2, 7). So you cannot expect them to combine in any way that will give you fewer than four terms in the answer.