Solve the inequalities?

Question

x^2+x+1>0

Answer 1

This inequality is true for all x.
To show this, complete the square:
x^2+x+1
= (x+1/2)^2 + 3/4
The latter expression is always positive.

Or find the discriminant:
b^2-4ac
= 1-4(1)(1)
= -3 < 0
Since discriminant <0 and coefficient of x^2 is positive, the quadratic trinomial is always positive.

Answer 2

Look at the discriminant D = b^2 - 4 ac = 1^2 - 4(1)(1) = -3. When the D < 0, the roots are not real. This means that the graph never crosses the x-axis. Since f(0) = 1 is positive, the function is positive for all x. That is, x^2 + x + 1 > 0 for all x.

Answer 3

You can plot it or you can take a derivative to find the minimum point.

f = x^2 + x + 1
df/dx = 2x + 1 = 0 and x = -1/2
f = 3/4 so the minimum of the function is greater than 0 so the statement is true for all values of x

d^2f/dx^2 = 2 so x=-1/2 is a minimum