[SPLIT] product of radical expression and its conjugate

daniel

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Nov 26, 2007
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15
how about
what is the product of 2 radical 7 + 3 radical 5 and its conjugate?
 
Re: radical expressions

Well what do you think - show some work.

Hint:

(a+b) (ab)=a2b2\displaystyle (a + b) \cdot\ (a - b) = a^2 - b^2
 
Re: radical expressions

would this be it?
2 rad.7 + 3 rad.5 = 2 rad. 7 - 3 rad.5
(2+3) 7exp.2 -5 exp. 2 = (7 +5) (7-5)

I pretty sure this is wrong. I have been looking over my class notes and my algebra book, it seems to say that I cannot add or subtract unlike radicand numbers. If that is so, why would my teacher have on my assignment such problem. My text went as far as stating that it would not touch on unlike radicands.
also, how do you put symbols in response?
thanks
 
Re: radical expressions

You don't equate them together.

You are given:
27+35\displaystyle 2\sqrt{7} + 3\sqrt{5}

Its conjugate is:
2735\displaystyle 2\sqrt{7} - 3\sqrt{5}

You are told to multiply them:
(27+35)(2735)\displaystyle \left(2\sqrt{7} + 3\sqrt{5}\right) \cdot \left(2\sqrt{7} - 3\sqrt{5}\right)

Do you know how to multiply binomials?

Also, if you are familiar with the difference of squares as shown in Subhotosh Khan's post, then imagine:
a=27\displaystyle a = 2\sqrt{7} and b=35\displaystyle b = 3\sqrt{5}.
 
4(7)29(5)2\displaystyle 4 (\sqrt{7})^{2} - 9 (\sqrt{5})^{2}

Yep! You can simplify it even more though. What are (7)2\displaystyle (\sqrt{7})^{2} and (5)2\displaystyle (\sqrt{5})^{2} equal to?
 
Not quite. You're multiplying 4 & 7 as well as 9 & 5, not adding them.
 
Yep, correct! You can always verify with your calculator, too!
 
WOW, THANKS! I struggled with this assignment. then when someone goes over and explains, it seems so much easier.
I want to be able to understand it and trying not to rely on my calculator as much, but I did see how I could verify answer once I did work first. Thank you, you have no idea how much you have helped me tonight. Again thanks for tutoring me through every step. I appreciate very much. Have a good night!
 
Sorry I have just one more question
3a squared -1=11

then do i subtract 11 from both sides and get 3a squared -1 -11=0
 
Yep that works. And when you subtract -11 from -1, you will end up with:
3a212=0\displaystyle 3a^{2} - 12 = 0

This should tell you that you're dealing with a difference of squares but just a hint, I would factor out something before applying the difference of squares formula :wink:
 
I'm guessing you made a typo when you wrote 3 = 0 o_O
But yes, a = 2 and -2 :wink:
 
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