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My teacher gave me a question which I am not able to solve:

If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$

I tried to substitute the value of x in the expression, but that comes out to be very big.

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  • $\begingroup$ What do you mean by "very big" ? $\endgroup$ – lmsteffan Jun 19 '15 at 13:30
  • $\begingroup$ If $x=2+\sqrt3$ then $x-2=\sqrt3$ hence $(x-2)^2=3$, that is, $x^2-4x+1=0$, that is, $x+\frac1x=4$ hence $\left(x+\frac1x\right)^2=16$, that is, $x^2+2+\frac1{x^2}=16$ hence $x^2+\frac1{x^2}=14$. $\endgroup$ – Did Jan 23 '17 at 22:38
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Here is a slightly weird way of doing it. $x$ looks like the quadratic formula, so if we can cook up a quadratic equation that it satisfies, we won't actually have to square it. The solutions to $$ y^2+2by+c = 0 $$ are $$ y = -b \pm \sqrt{b^2-c} $$ (because $a=1$ and the $b$ has a $2$ multiplying it that cancels the $2$ normally present). In this case, the number not in the square root is $2$, so $b=-2$. Then $b^2=4$, so to agree with the number inside the square root, we have $ 4-c=3 $, so $c=1$. Therefore $x$ satisfies the quadratic equation $$ x^2 -4x+1 = 0 \tag{1} $$ From here, we can read off that $$ x^2 = 4x-1, $$ and if we divide (1) by $x$, we also find $$ \frac{1}{x} = 4-x, \tag{2} $$ and dividing (1) by $x^2$ gives $$ \frac{1}{x^2} = \frac{4}{x}-1 = 4(4-x)-1 = 15-4x, $$ applying (2) to the $1/x$ term. Therefore, $$ x^2 + \frac{1}{x^2} = 4x-1 + 15-4x = 14. $$

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HINT :

$$\frac{1}{x}=\frac{1}{2+\sqrt 3}=\frac{2-\sqrt 3}{(2+\sqrt 3)(2-\sqrt 3)}=2-\sqrt 3$$

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Hint: Try to find a simple expression for $1/x$.

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    $\begingroup$ I don't think this should be posted as an answer, it stands to reason that this very simplification is problematic for the OP, so you should elaborate a bit. $\endgroup$ – Victor Jun 19 '15 at 13:32
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    $\begingroup$ @Victor: When in doubt I prefer to reveal as little as possible, not as much as possible. From the question it's not clear that the OP has focused on isolating the $1/x$ part, and perhaps merely knowing where to look is all he needs. (I gather he has already seen examples of this simplification, otherwise he wouldn't have been asked this question.) $\endgroup$ – Meni Rosenfeld Jun 19 '15 at 13:35
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    $\begingroup$ @Victor: Between mathlove's and Essam's answers, the OP now has a complete solution to his homework question. I'm new here, but I doubt this is what this site is about... $\endgroup$ – Meni Rosenfeld Jun 19 '15 at 13:39
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    $\begingroup$ @G.Sassatelli: I maintain my position that merely being told where to look (so that he knows he should keep looking there until something is found) might just be all the OP needed. I will keep your analysis in mind, though. $\endgroup$ – Meni Rosenfeld Jun 19 '15 at 14:19
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    $\begingroup$ @George: Hm? This is obviously not a critique or clarification request. I provided a hint (a subtle one, but still), under the assumption we should not just give complete answers to homework questions. $\endgroup$ – Meni Rosenfeld Jun 19 '15 at 14:21
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Hint: $$(2-\sqrt{3})(2+\sqrt{3})=1$$ $$(2-\sqrt{3})(x)=1$$ so $$\frac{1}{x}=2-\sqrt{3}$$ $$(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$$ $$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2$$ $$x^2+\frac{1}{x^2}=(2+\sqrt{3}+2-\sqrt{3})^2-2=14$$

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Since everyone else has answered with various shortcut methods, let me just show you how you could simply substitute in directly and get an answer cleanly without things ever getting 'too big'.

We know that $x=2+\sqrt{3}$, so we can square this using the usual $a^2+2ab+b^2$ binomial formula: $x^2=2^2+2(2)(\sqrt3)+(\sqrt3)^2$ $= 4+4\sqrt3+3$ $=7+4\sqrt{3}$.

Now, $\dfrac1{x^2}$ is $\dfrac1{7+4\sqrt3}$; we can use the usual method for rationalizing the denominator to handle this, by multiplying by $\dfrac{7-4\sqrt3}{7-4\sqrt3}$. This gives $\dfrac1{x^2}$ $=\dfrac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$ $=\dfrac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$ $=\dfrac{7-4\sqrt3}{7^2-4^2\cdot 3}$ $=\dfrac{7-4\sqrt3}{49-48}$ $=7-4\sqrt3$.

The most important lesson here is that whenever you have an expression of the form $x=a+b\sqrt{c}$, then the powers of $x$ — both positive and negative — will never get too messy; they'll all always be of the form $p+q\sqrt{c}$ for some rational $p$ and $q$. In this case, $p$ and $q$ were integers even for the negative powers of $x$; that won't always be the case, but they'll always be rational.

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Siddhant, when you said that you tried to substitute the value of $x$, I could understand why your calculations went too far, and what you meant by 'very big'.

But here's a very short and easy way of solving your question. Write $x^2 + \frac {1}{x^2} = (x+\frac1x)^2 - 2$ and then solve.

$$\implies 2+\sqrt3+ \frac {1}{2+\sqrt3} = \frac {(2+\sqrt3)^2 + 1}{2+\sqrt3} $$ $=4$

Now, what remains of your question is $4^2 - 2$ which is $14$.

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  • $\begingroup$ thnx a lot for the help! $\endgroup$ – codetalker Feb 20 '17 at 15:33
  • $\begingroup$ Welcome Siddhant, and thanks for upvoting ! $\endgroup$ – Saksham Feb 20 '17 at 19:16
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By brute-force:

$$(2+\sqrt3)^2+\frac1{(2+\sqrt3)^2}=\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}.$$

Then

$$(2+\sqrt3)^2=7+4\sqrt3,\\ (2+\sqrt3)^4=(7+4\sqrt3)^2=97+56\sqrt3,$$

$$\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}=\frac{98+56\sqrt3}{7+4\sqrt3}=\frac{14\cdot7+14\cdot4\sqrt3}{7+4\sqrt3}=14.$$

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By the Vieta formulas, $2\pm\sqrt3$ are the roots of the quadratic equation $x^2-4x+1=0$, or $$x+\frac1x=4.$$ Squaring

$$x^2+2+\frac1{x^2}=16.$$

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