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$$\sqrt{x^2+x+1}=1-x-x^2$$ I am doing this problem for about an hour and I cant get to all the results. What is the easiest way to do this? I tried by squaring everything right away but at the I end I get a pretty messed up equation that is not easy to factor at all: $x(-x^3-2x^2+2x+3)=0$. Of course I see that one answer is $x=0$ but how do I get other answers?

RiktasMath
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    Hint: Can you solve $\sqrt{1+u}=1-u$? – Did Apr 07 '18 at 13:33
  • Cautionary note - when squaring an equation like this you may get extra solutions, so you always need to substitute back into the original equation. (This is because if you square $y=a$ to obtain $y^2=a^2$, the second equation is also satisfied by $y=-a$) – Mark Bennet Apr 07 '18 at 13:41
  • @RiktasMath Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 10 '18 at 12:20

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Hint

$$\sqrt{x^2+x+1}=1-x-x^2$$ $$\sqrt{x^2+x+1}=2-(1+x+x^2)$$ $$\sqrt{u}=2-u$$ Square it ...

user577215664
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Note that

$$x(-x^3-2x^2+2x+3)=-x(x+1)(x^2+x-3)$$

user
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  • How was I supposed to see that? – RiktasMath Apr 07 '18 at 13:34
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    By rational root theorem https://en.wikipedia.org/wiki/Rational_root_theorem a root for $(-x^3-2x^2+2x+3)$ must be among 3,1,-1,-3 – user Apr 07 '18 at 13:37
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    It must be one of those 4 numbers if it's rational, i.e., if the polynomial can be factored over the integers. – G Tony Jacobs Apr 07 '18 at 13:38
  • @Did Those as already G Tony Jacobs observed it must be if it is rational, and those are the values we can use to find a factorization that indeed works for x=-1. – user Apr 07 '18 at 14:04
  • Yeah. Hence, yet again a wrong comment. – Did Apr 07 '18 at 14:04
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    Whatever the precise merits of the comment, using the rational root theorem to find an easy factor of a cubic (if it exists) is a try worth noting for OP. – Mark Bennet Apr 07 '18 at 14:06
  • @MarkBennet Of course. And overlooking that this criterion "misses" the non rational roots is seriously misleading the OP. – Did Apr 08 '18 at 12:24
  • ((Yet again a factual comment addressing a mathematical error, which "conveniently" disappears.)) – Did Apr 08 '18 at 12:24
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For $x \ge 0$, note that $\sqrt{x^2 + x + 1}$ gets bigger when $x$ increases while $1 - x - x^2$ get's smaller. Since an upward sloping line and a downward sloping line cross in only one place, the only solution is $x = 0$.

For $x \le -1$ the same reasoning shows $-1$ is the only solution, so it only remains to check in between $-1$ and $0$.

Since $f(x) = x^2 + x + 1$ is "curved upwards" (as $x^2$ has a positive coefficient), and $f(-1) = f(0) = 1$, we know $f(x)$ is less than $1$ in between $-1$ and $0$. Since $\sqrt f < 1 \iff f < 1$, it follows that $\sqrt{x^2 + x + 1}$ < 1 when $-1 < x < 0$.

Since $g(x) = 1 - x - x^2$ is "curved downwards" and $g(-1) = g(0) = 1$, we have $1 - x - x^2 > 1$ when $-1<x<0$. So the only two solutions are $-1$ and $0$.

Suhas Vijaykumar
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