Solve equation: $\sqrt{x^2+x+1}=1-x-x^2$

Question

$$\sqrt{x^2+x+1}=1-x-x^2$$ 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?

Answer 1

Note that

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

Answer 2

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 ...

Answer 3

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 $0$.