Answer to Question #13598 in Algebra for jenthura

Question #13598
My problem is conjugates. Given (sqrt2-5)/(sqrt2-2) Obviously, the conjugate of (sqrt2-2) is (sqrt+2), however, (sqrt2-2) is equal to (-2+sqrt2), correct? All I did was move the sqrt2 to the other side of -2. I changed no signs, so it must still be equal. Since it is still equal, the new conjugate, (-2-sqrt2), should still work, correct? No. It doesn't. Normally with my algebra book (Saxon Algebra 2) I work the problem as given and it comes out fine. However, sometimes I work it as given and the answer is wrong. In the problem I gave you, the answer (as stated in the correction book)* is (-6-5sqrt2). That answer is only possible when I change the denominator as I did above (sqrt2-2 --> -2+sqrt2). Is it my book's fault? If so, then Saxon has major problems, because I have encountered this same problem several times. *I have access to the correction book because I'm homeschooled.
Expert's answer
Firstly let's work your problem out:
=(-8-3sqrt2)/{-2}=4+1.5 sqrt2 that's the answer

Now let's try the second way you described (we'll the same answer, of course):
& (sqrt2-5)/(sqrt2-2)=(sqrt2-5)/(-2+sqrt2)=(sqrt2-5)(-2-sqrt2)/{(-2+sqrt2)(-2-sqrt2)}=
=-(sqrt2-5)(2+sqrt2)/{-(-2+sqrt2)(2+sqrt2)}=(sqrt2-5)(2+sqrt2)/{(-2+sqrt2)(2+sqrt2)} so we've obtained the same thing and the answer is the same, of course. The new conjugate, as you call it, still works. If you change no signs expression must remain the same. The point of conjugates in this task and alike is to remove radicals from the denominator. For that purpose we're forming product of the kind (a-b)(a+b)=a^2-b^2& in the denominator. Obviously, order of terms with or without radicals (sqrt signs) doesn't matter

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