Spivak Calculus 3rd Edition Chapter 1 Problem 4 (vi)

Question

I'm not sure about arriving to the solution for $x^2+x+1>2$
I've gotten as far as:
$x^2 + x + 1 > 2$
$x(x + 1) + 1 > 2$
$x(x + 1) + 1 + x > 2 + x$
$x(x + 1) + 1(x + 1) > x + 1 + 1$
$(x + 1)^2 > (x + 1) + 1$
$(x + 1) > 1 + (x+1)^{-1}$
$(x + 1) - (x+1)^{-1} > 1$
$x + 1 - \frac{1}{(x+1)} > 1$
$x - \frac{1}{(x+1)} > 0$

but the actual solutions is:
$x > \frac{[-1 + \sqrt5]}{2}$
or
$x < \frac{[-1 - \sqrt5]}{2}$

Can someone advise on this one, thanks?

Answer 1

It's $$x^2+x+\frac{1}{4}>\frac{5}{4}$$ or $$\left(x+\frac{1}{2}\right)^2>\frac{5}{4},$$ which gives the answer: $$x>\frac{\sqrt5-1}{2}$$ or $$x<\frac{-1-\sqrt5}{2}.$$

Answer 2

you have to solve $$x^2+x-1>0$$ this is equivalent to $$\left(x+\frac{1}{2}\right)^2>\frac{5}{4}$$ and it remaines to solve $$|x+1/2|>\frac{\sqrt{5}}{2}$$ the solution is given by $$x<\frac{1}{2}(-1-\sqrt{5})$$ or $$x>\frac{1}{2}(-1+\sqrt{5})$$

Answer 3

$(x+1/2)^2 -1/4 +1 -2 >0.$

$(x+1/2)^2 -5/4 >0.$

$(x+(1/2) -\sqrt{5/4})\cdot$

$(x+(1/2) +\sqrt{5/4}) >0.$

1) Both factors are $>0$.

$x+(1/2) - (1/2)√5 >0$, or

$x> (1/2)√5 -(1/2)$.

2) Both factors are $ <0.$

$x+(1/2) + (1/2)√5 <0$, or

$x < - (1/2)√5 -1/2.$

Used : $a^2-b^2 =(a-b)(a+b).$

Where it says both factors are positive, resp. negative only one factor is given, why ?