Question

x^2+x>1

Answer 1

one way is to factor x^2+x-1 =0 into
(x-a)(x-b)=0
then argue we want
(x-a)(x-b)>0 requires either (1) both (x-a) and (x-b) are positive
or
(2) both terms are negative.

Answer 2

I would use the quadratic formula to find
\[ x= \frac{-1 \pm \sqrt{5}}{2} \]
so that we have
\[ (x +\frac{1-\sqrt{5}}{2})(x +\frac{1+\sqrt{5}}{2})>0 \]
can you finish ?

Answer 3

\[x^2+x>1 => x^2+x+(1/4)-(1/4)>1 => (x+(1/2)^2)>(3/4) => x+(1/2)>\sqrt{3/4} => x>(\sqrt{3}-1)/2\]

Answer 4

@parto I cant see all of your solution.

Answer 5

ooh sorry ... wait i type again

Answer 6

\[x^2+x>1 => x^2+x+(1/4)-(1/4)>1 \]

Answer 7

\[(x+(1/2))^2>3/4 => x+(1/2)>\sqrt{(3/4)}\]

Answer 8

\[x>\sqrt{3/4}-(1/2) => x>(\sqrt{3}-1)/2\]

Answer 9

ok ? got it ?

Answer 10

soory its \[(\sqrt{5}-1)/2\]

Answer 11

@parto you are missing the other solution.