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$$x+\frac{1}{x}=25 + \frac{1}{25}$$

The solution is very simple. But the problem is whether my solution is correct or not. I did it by simply comparing the LHS and the RHS. Thus, I got $x=25$ or $\frac{1}{25}$. But my book does it in this way

$x-25=\frac{1}{25}-\frac{1}{x}=\frac{x-25}{25x}$. So, $x=25$ or $1=\frac {1}{25x}\implies x=\frac{1}{25}$.

I asked my teacher whether my method was correct or not. She told me that the method in the book is correct and that my method of comparing will not be accepted during the exam as it is 'not mathematical' and is 'some sort of hit and trial'.

Now, I am not worried about whether I'll be awarded marks for my method or not. But is it 'not mathematical' to compare the L.H.S. and R.H.S in such type of equations?

Leucippus
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Soham
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  • Both $\frac{1}{25}$ and $25$ are right here. – Taylor Oct 25 '16 at 16:59
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    It depends on the context, I guess you're asked about quadratic equations.
    Say to solve $x$ in $$(x – a)(x – b) = (y – a)(y – b)$$ you may get the trivial solution $x=y$ at once. However, another solution is not straight forward, you need to further work it out. Sometimes trial-and-error is necessary, for example when applying factor theorem. In your case, it's better to rewriting into a quadratic equation and then solve for it.     – Ng Chung Tak Oct 25 '16 at 17:12

2 Answers2

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By comparing the LHS and RHS, you found a few solutions.

The question is have you found all the solutions?

If you can justify that you have found all the solutions, then I don't see anything wrong with hit and trial.

Do not confuse schooling with educations, but there is no point going against the grading system in school.

Siong Thye Goh
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  • Yeah! I'm not worried about that. But can it be said that as a quadratic polynomial cannot have more than two roots, I have found all the roots? – Soham Oct 25 '16 at 17:00
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    If you multiply by $x$, indeed you have converted the question to a quadratic equation problem and it has at most $2$ roots. We found $2$ numbers that satisfy the equation and we have completely solve the problem. – Siong Thye Goh Oct 25 '16 at 17:04
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    If you multiply by $x$ you may introduce the solution $x=0$. You have to be careful doing these type of manipulations. – LinAlg Oct 25 '16 at 17:06
  • Yup, we have to verify that doesn't happen. – Siong Thye Goh Oct 25 '16 at 17:07
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The "longer" method is: \begin{align} x + \frac{1}{x} &= 25 + \frac{1}{25} \\ x^{2} + 1 &= \left(25 + \frac{1}{25} \right) x \\ x^{2} - \left(25 + \frac{1}{25} \right) x + 1 &= 0 \\ x &= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \sqrt{\left(25 + \frac{1}{25} \right)^2 - 4} \\ &= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \sqrt{\left(25 - \frac{1}{25} \right)^2} \\ &= \frac{1}{2} \, \left(25 + \frac{1}{25} \right) \pm \frac{1}{2} \, \left(25 - \frac{1}{25} \right) \\ x &= 25 \hspace{5mm} \mbox{or} \hspace{5mm} \frac{1}{25}. \end{align}

Leucippus
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    Alternatively, factorize $$x^2-\left( 25+\frac{1}{25}\right)x+1=\left( x-25\right)\left(x-\frac{1}{25} \right)$$ – Siong Thye Goh Oct 25 '16 at 17:11
  • I had a number-theory teacher who defined mathematics as the art of being lazy. I'm not entirely happy with that definition, but there is a germ of truth in it. My teacher would have been delighted with your proof. Anyway it is a proof. Note that the logic of the longer proofs is If LHS =RHS, then x=25 or x=1/25. Having winnowed out the only possible answers, the next step is to do just what you did: see that these numbers work. Of course, being lazy, we do this only once when we derive the quadratic formula and then check that the answers it supplies do indeed work. – Airymouse Oct 25 '16 at 17:31
  • One more thing, your method is certainly not some sort of "hit or trial." You were not guessing, you were thinking. The notion that you just guessed that 25 and 1/25 were the two solutions is absurd. – Airymouse Oct 25 '16 at 20:48
  • @Airymouse Yes, this was not "a reasonable guess" method presented and was indicated in the "longer" method phrase. In the question proposed the "reasonable guess" approach can be viewed in several ways, as has been demonstrated. The method I have presented would also cover equations of the type $x + \frac{1}{x} = \sqrt{3} + \frac{1}{\sqrt{2}}$ whereas "a reasonable guess" method would not. – Leucippus Oct 25 '16 at 21:38