Rationalizing denominators in radical expressions

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MickieMik

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Rationalize the denominator

(2)/ sq rt [3] + sq rt [2]

I know this probably seems easy to some but from what I can find about how to do a problem of this type you are supposed to multiply the denominator by both the numerator and the denominator. So would you do this?

Numerator (2)(sq rt 3) + (2)(sq rt 3) = 6+6 = 12
Denominator (sq rt 3) + (sq rt 2) (sq rt 3) +(sq rt 2) = (sq rt 9) + (sq rt 4)

(12)/(3 + 2) = 12/5 (?) I have no idea of how to do this or if I am even close. This just looks kind of crazy , Please help. Thank you very much
 
(2)/ sq rt [3] + sq rt [2] means...
\(\displaystyle \frac{2}{\sqrt{3}} + \sqrt{2}\)

Is this what you mean or do you mean 2/(sq rt [3] + sq rt [2])?

If it is the latter, you need to multiply both numerator and denominator by the conjugate of the denominator. That should get you started.
 
You must get the idea of the "congugate" based on the concept \(\displaystyle (x+y)(x-y) = x^{2} - y^{2}\).

So,

\(\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}}*\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)

Now multiply and observe the workings of the concept mentioned above.
 
Loren, thanks for responding. The problem is written like a fraction with 2 as the numerator and the denominator written as ?[6] + ?[2]

Thanks
 
MickieMik said:
Loren, thanks for responding. The problem is written like a fraction with 2 as the numerator and the denominator written as ?[6] + ?[2]

Thanks
Click to expand...
In other words, exactly as I described and demonstrated. Let's see what you get.
 
I really don't get this from other questions/answers in the book but this is a bad attempt.

2/ sq rt[3] + sq rt [2]
sq rt [6] - sq rt [4] / 9 - 4
2 sq rt[6] / 4

From what I could see in my book this is what I came up with. Also in my book and also CD It looks to me when simplifying a radical when there is exponents in it you do this for example

sq rt[9x^12y^8x^10 you find the square of the 9 and then to me it looks like when squaring you do this

3x^6y^4z^10
3x^3y^2z^5. In the book or CD it doesn't get as high as 10 for an exponent (squaring it) but say an x^7 becomes x^6 and that becomes x^3, so if we simplify an even number we take the largest even number out of an exponent and then simplify it again by dividing in half again, cubed by 3 and 3 again if possible? Of course if there was an odd exponent for squared, even for cubed, etc you would leave that one behind.
 
Rationalizing the denominator in Radicals

I really don't get this from other questions/answers in the book but this is a bad attempt.

2/ sq rt[3] + sq rt [2]
sq rt [6] - sq rt [4] / 9 - 4
2 sq rt[6] / 4

From what I could see in my book this is what I came up with. Also in my book and also CD It looks to me when simplifying a radical when there is exponents in it you do this for example

sq rt[9x^12y^8x^10 you find the square of the 9 and then to me it looks like when squaring you do this

3x^6y^4z^10
3x^3y^2z^5. In the book or CD it doesn't get as high as 10 for an exponent (squaring it) but say an x^7 becomes x^6 and that becomes x^3, so if we simplify an even number we take the largest even number out of an exponent and then simplify it again by dividing in half again, cubed by 3 and 3 again if possible? Of course if there was an odd exponent for squared, even for cubed, etc you would leave that one behind.
 
Re: Rationalizing the denominator in Radicals

Your statement 2/ sq rt[3] + sq rt [2] means...

\(\displaystyle \frac{2}{\sqrt{3}}+\sqrt{2}\)

Is that what you mean? If not, please clarify by using parenthesis so that there is no misunderstanding.
 
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