I was trying to find if there a method similar to multiplying and dividing by the conjugate $$\frac{1}{\sqrt{3x+5}-\sqrt{5x+11} - \sqrt{x+9}},$$ but that doesn't seem to work here. Also, is there a method of multiplying by a conjugate for roots other than the square root? Such as $(ax+b)^\frac1n ± (cx+d)^\frac1m$

  • $\begingroup$ The second question is more generic, it may depend on the (possible) factorisation of $p^n \pm q^m$. $\endgroup$ – Macavity Sep 17 '14 at 8:41
  • $\begingroup$ say if I had a fifth root of a polynomial and a cube root of another polynomial, how would I find if this qualifies for conjugation ? $\endgroup$ – user142795 Sep 17 '14 at 8:42
  • $\begingroup$ $p^5 \pm q^3$ does not factorise in general :(. $\endgroup$ – Macavity Sep 17 '14 at 8:44
  • $\begingroup$ I see, so how could I do the such with a trinomial of square roots? $\endgroup$ – user142795 Sep 17 '14 at 8:46
  • $\begingroup$ Already shown below. $\endgroup$ – Macavity Sep 17 '14 at 8:46

In case you want to approach the general one, if you see the following in denominator, $$(p^{1/n}-q^{1/m})$$ First focus on p $$(p^{1/n}-q^{1/m})=(p^{1/n}-(q^{n/m})^{1/n})$$ multiply that by $p^{(n-1)/n}+p^{(n-2)/n}q^{n/m}+...+(q^{n/m})^{(n-1)/n}$
You get $$p-q^{n/m}=(p^m)^{1/m}-(q^n)^{1/m}$$ then again multiply this by $(p^m)^{(m-1)/m}+...(q^n)^{(m-1)/m}$

you'll get $p^m-q^n$

  • $\begingroup$ How is it that you have derived this? Is this a relatively original idea? $\endgroup$ – user142795 Sep 17 '14 at 10:10
  • $\begingroup$ Nice.... Now for the plus sign... $ +1$ $\endgroup$ – Macavity Sep 17 '14 at 10:19
  • $\begingroup$ Yes, would this method still work with the sum ? $\endgroup$ – user142795 Sep 18 '14 at 10:26
  • $\begingroup$ You would just change the sum into a difference when multiplying by the various powers of p and q, correct? $\endgroup$ – user142795 Sep 18 '14 at 10:29

You can apply it twice: $$\frac1{\sqrt a-\sqrt b-\sqrt c}=\frac{(\sqrt a-\sqrt b)+\sqrt c}{(\sqrt a-\sqrt b)^2-c}=\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}},$$ then $$\frac{\sqrt a-\sqrt b+\sqrt c}{a+b-c-2\sqrt{ab}}=\frac{(\sqrt a-\sqrt b+\sqrt c)(a+b-c+2\sqrt{ab})}{(a+b-c)^2-4ab}.$$


Hint: $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right) \left(\sqrt{a}-\sqrt{b}-\sqrt{c}\right) \left(-\sqrt{a}+\sqrt{b}-\sqrt{c}\right) \left(-\sqrt{a}-\sqrt{b}+\sqrt{c}\right) \\ = a^2+b^2+c^2-2(ab+bc+ca)$$


Try this for size:

For the second part of your question, see math.stackexchange.com/questions/838287 .


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