x^2 + x +1 =0

This function will always have a positive value and there us no x value in which f(x)=0

To verify, try to find the function's roots:

x1= 1+(sqrt(1-4)= 1-sqrt(-3)

x2= 1-sqrt(-3)

which is impossible because the square root can not contain negative values for real numbers.

Then the function has no solution. Or in other words the function does not intersect with y-axis.

To find the f(2^x) = 0 where f(x) =x^2+x+1.

Solution:

Let F(2^x) = 0, gives y^2+y+1 = 0 where y = 2^x.

The solution of y is given by :

y1 = (-1+sqrt(1-4))/2 = )-1+sqrt(-3))/2 and

y2 = (-1-sqrt(-3))/2.

So 2^x = (-1+sqrt(-3))/2 = { cos (2pi)/3 +isin(2pi/3)}

2^x = e^i(2pi/3). Or

x = i(2pi/3)/log2 which is an imaginary .

Similarly

2^x = (-1/2-sqrt(-3))/2 = cos(4pi/3) +i sin(4pi)/3 = e^i(4pi/3). Or

x = i(4pi)/3]/log2 which is imaginary.

So this has only imaginary solution.

x = e^i(4pi)/3

Let's recall the rule of an exponential function, namely that the values of the function are always positive and never zero.

f(x) = a^x, where a>0 and x is a real number.

f:R->(0,+inf.)

We notice that that the resulted expression (2^x)^2 + 2^x + 1 will always be positive and it will not be equal to 0, for any value of x from the R set.