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My goal was to remove the radical from the numerator of the function: $$f(x)=\frac{\sqrt{x^2+9}-3}{x^2}$$ I first entered my function.

f[x_] = (Sqrt[9 + x^2] - 3)/x^2 

Then I created a list.

lst = {Numerator[f[x]], Denominator[f[x]]}

Then I multiplied both components of the list by $\sqrt{x^2+9}+3$, simplifying the result.

lst = lst*(Sqrt[9 + x^2] + 3) // Simplify

Then I changed the list back to a fraction.

f[x_] = lst /. {x_, y_} -> x/y

Which gave me: $$\frac{1}{3+\sqrt{9+x^2}}$$

I have a couple of questions.

  1. Is there a simpler way to convert my last list back to a fraction?

  2. Anyone have a simpler overall process for this particular sequence of algebraic manipulations?

  3. I know I've seen a page on Mathematica Stack Exchange where there is a long list of how to simplify and change algebraic expressions, but I've been unable to find it. If someone knows of these links, can they share them?

Thanks.

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    $\begingroup$ Divide @@ lst should work for you. $\endgroup$ – J. M. will be back soon Dec 17 '15 at 1:13
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    $\begingroup$ Simplify[f[x]*(Sqrt[9 + x^2] + 3)]/(Sqrt[9 + x^2] + 3) removes the need for the Divide $\endgroup$ – bill s Dec 17 '15 at 1:46
  • $\begingroup$ @David, why not just FullSimplify[f[x]] for the initial statement? $\endgroup$ – garej Dec 19 '15 at 20:05
  • $\begingroup$ @garej Because it is not just the answer that I am interested in helping my students with. Yes, I will show them the Simplify and the FullSimplify command to help them check their answers. But there is a lot of detailed hand work to do in evaluating limits and derivatives and if they don't get the correct answer, I want to be able to show them how they can check some of their step-by-step work in order to discover where they have made their error. $\endgroup$ – David Dec 20 '15 at 20:29
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Simplification is in the eye of the beholder. One approach is

f = (Sqrt[9 + x^2] - 3)/x^2;
1/FullSimplify[1/f]
(* 1/(3 + Sqrt[9 + x^2]) *)
FullSimplify[x^2 %]/x^2
(* (-3 + Sqrt[9 + x^2])/x^2 *)

You may be thinking of this question.

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