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Solve for $x$ and $n$ in the equation $$ \dfrac{x^2 + 7x +1}{ x^2 + x +1}= 3n,\qquad n \in \Bbb{Z}.$$ The original problem was a trigonometric equation but solved it till I got stuck here. I am a high school student who is self studying and preparing for college entrance exam. Original problem was a trigonometric equation which came from 'cengage exam crack'.

[After Mark's and Jyrki's hints in the comments, I got the answer and posted it below.] Original problem: enter image description here

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  • $\begingroup$ Is the value of $n$ known, or also for you to figure out? $\endgroup$ Apr 12, 2019 at 3:33
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    $\begingroup$ Have you tried clearing fractions and using the quadratic formula? You need the bit under the square root to be positive. @JyrkiLahtonen's comment gives you the possible values of $n$ $\endgroup$ Apr 12, 2019 at 3:44
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    $\begingroup$ @MarkBennet thank you. I got the answer. I also posted in below. $\endgroup$
    – user541396
    Apr 12, 2019 at 4:07
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    $\begingroup$ While I have nominated this question for reopening, I would personally like to seem more context included in the question. Specifically, you say that the problem started life as a trigonometric equation. Editing this original trigonometric equation into the question would help pin down questions of domain (i.e. what values of $x$ and $n$ actually make sense, given that they have to be fed to some trig function). A more complete reference to "engage exam crack" (whatever that is) would also be helpful. $\endgroup$
    – Xander Henderson
    Apr 15, 2019 at 16:41
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    $\begingroup$ @XanderHenderson what do you mean by 'more complete reference'? $\endgroup$
    – user541396
    Apr 17, 2019 at 13:42

2 Answers 2

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1) Convert the problem into quadratic $ x^2(1-3n) + x(7-3n) + 1-3n = 0 $

2) set discriminant greater than equal to $0$

3) find the range of $n $

4) find the possible values of n which are $(0,1,-1)$

5) plug them back in to find values of $x$

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  • $\begingroup$ That's a nice way of combining the steps. Well done! $\endgroup$ Apr 12, 2019 at 6:40
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$3n =\dfrac{x^2 + 7x +1}{ x^2 + x +1} $ so $3nx^2+3nx+3n =x^2+7x+1 $ so $(3n-1)x^2+(3n-7)x+3n-1 =0 $ or

$\begin{array}\\ x &=\dfrac{-3n+7\pm \sqrt{(3n-7)^2-4(3n-1)^2}}{2(3n-1)}\\ &=\dfrac{-3n+7\pm \sqrt{(3n-7-2(3n-1))(3n-7+2(3n-1))}}{2(3n-1)}\\ &=\dfrac{-3n+7\pm \sqrt{(-3n-5)(9n-9)}}{2(3n-1)}\\ &=\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)}\\ \end{array} $

If nothing is specified about $x$, this is as far as we can go.

If $x$ has to be real, then $(3n+5)(n-1) \le 0$ so that $-\dfrac53 \le n \le 1$. Since $n$ is an integer, $n = -1, 0, 1$.

If $x$ is supposed to be rational, then $(3n-7)^2-4(3n-1)^2 =m^2 $. From the preceding result, $n=1 \implies m^2 =4^2-4(2)^2 =0 $, $n=0 \implies m^2 =7^2-4(1)^2 =45 $, $n=-1 \implies m^2 =10^2-4(-4)^2 =36 $.

Therefore the only rational solutions with integer $n$ are $n=1, 0, -1$ for which $n=1 \implies x =\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)} =\dfrac{4}{4} =1 $, $n=0 \implies x =\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)} =\dfrac{7\pm 3\sqrt{5}}{-2} $, and $n=-1 \implies x =\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)} =\dfrac{-10\pm 4}{-8} =\dfrac{-14, -6}{-8} =\dfrac74, \dfrac52 $.

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    $\begingroup$ Why did you exclude n= 0, if it is because it doesn't provide a rational solution , then question doesn't say anything about X to be rational or irrational but only real. $\endgroup$
    – user541396
    Apr 12, 2019 at 5:16
  • $\begingroup$ You are right. Added that case and upvoted your comment. Thanks, $\endgroup$ Apr 12, 2019 at 5:26
  • $\begingroup$ The question doesn't actually say anything about $x$ being real, so why is everyone making that assumption? $\endgroup$ Apr 15, 2019 at 5:45

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