Short Inequalities

1) x(x+1)>-1? ----> x^2+x+1>0 /solve as equation and find critical points where x=0, then plug in and assign the sign-decide on inequality condition(s)/
x^2+x+1=0, has no solution as D is -ve.

2) x(x^2+x+1)>-1? ----> x^3+x^2+x+1>0, (x^2 +1)(x+1)>0 /the same approach as above/ (x^2+1)(x+1)=0; x^2+1=0 OR x+1=0; x^2+1=0 ---> x doesn't have any solution as x^2 cannot be -ve and 1 is greater than 0, hence only possible x+1=0, x=-1. Plug in any value to the left from -1, x=-2 ---> (-2^2+1)(-2+1), hence x is -ve when x (-infinity;-1). Plug in any value to the right from -1, x=0 ---> (0^2+1)(0+1), hence x is +ve when x (-1; +infinity). Solution x<0 for -infinity<x<-1 and x>0 for -1<x<+infinity.

yellowho wrote:How do you solve for all value of X in these two short problems:

1) X(x+1)>-1?

2) x(x^2+x+1)>-1?

Without plugging in?

yellowho wrote:How do you solve for all value of X in these two short problems:

1) X(x+1)>-1?

2) x(x^2+x+1)>-1?

Without plugging in?

Sumbody tell me a conceptual approach to answer these question. I am highly confused. Thanks in advance

yellowho wrote:How do you solve for all value of X in these two short problems:

1) X(x+1)>-1?

2) x(x^2+x+1)>-1?

Without plugging in?

Hi there!

1) x(x+1) > -1 is equivalent to x^2 +x+1 > 0.

From the fact that the discriminant ("b^2-4ac") of the equation x^2 +x+1 = 0 is negative, we are sure that x^2+x+1 > 0 for all real values of x, and that´s what you were looking for.


2) x(x^2+x+1) > -1 is equivalent to x^3 +x^2 +x+1 > 0.

Please note that x^3 +x^2 +x+1 = x(x^2+1) + (x^2+1) = (x^2+1)(x+1) and from the fact that x^2 +1 >0 for all real values of x, then x^3 +x^2 +x+1 = (x^2+1)(x+1) > 0 if and only if x+1 > 0, that is, x > -1.

Regards,
Fabio.

yes, yellow the 1st stem has -D. I figured x(x+1)-1>0 above not what was originally posted. This one has no solution.
thanks Fabio, I just overlooked

fskilnik wrote:

yellowho wrote:How do you solve for all value of X in these two short problems:

1) X(x+1)>-1?

2) x(x^2+x+1)>-1?

Without plugging in?

Hi there!

1) x(x+1) > -1 is equivalent to x^2 +x+1 > 0.

From the fact that the discriminant ("b^2-4ac") of the equation x^2 +x+1 = 0 is negative, we are sure that x^2+x+1 > 0 for all real values of x, and that´s what you were looking for.


2) x(x^2+x+1) > -1 is equivalent to x^3 +x^2 +x+1 > 0.

Please note that x^3 +x^2 +x+1 = x(x^2+1) + (x^2+1) = (x^2+1)(x+1) and from the fact that x^2 +1 >0 for all real values of x, then x^3 +x^2 +x+1 = (x^2+1)(x+1) > 0 if and only if x+1 > 0, that is, x > -1.

Regards,
Fabio.