How does one show that x^2 + x + 1 > 0 for all values of x?


Complete the square:

[half the coefficient of the x term, decrease the power of the x^2 and x terms and put in brackets ^2, then correct the final term]

         x^2 + x + 1 = (x + 1/2)^2 + 3/4 > 0 for all values of x

since (x + 1/2)^2 > 0 and 3/4 > 0


Take the derivative of the function f(x) = x^2 + x + 1 and equate to zero:

[for each term, multiply the coefficient by the value of the power and decrease the power by one]

         d/dx (x^2 + x + 1) = 2x+1 = 0

Rearrange this to find x:

         => x = -1/2

Plug this into the function:

         f(-1/2) = (-1/2)^2 + (-1/2) + 1 = 1/4 – 1/2 + 1 = 3/4

-1/2 < 0 so the function has a minimum point at f(-1/2) (we know that this is a min not max since the function is a quadratic with positive x^2 coefficient). Since this is the only minimum (no. of extrema is one less than the highest power of x, which in this case is 2, so only one extremum) and this minimum is greater than zero, then the function must be greater than zero for all values of x.

         f(-1/2) = 3/4 > 0  =>  f(x) > 0 for all values of x


Determine the discriminant of the function f(x) = x^2 + x + 1

[The discriminant of a function f(x) = ax^2 + bx + c is given by b^2 – 4ac. Here a = 1, b = 1, c = 1]

         1^2 – 4*1*1 = -3

The discriminant is less than zero; therefore there are no real routes to the function (i.e. the function does not cross the x-axis). So the function has either a positive minimum or a negative maximum. Since the function is a quadratic with positive x^2 coefficient we know that it must have a positive minimum, therefore the function is positive for all x.

[b^2 – 4ac < 0 => complex conjugate pair of solutions;

b^2 – 4ac = 0 =>  one real (repeated) solution;

b^2 – 4ac > 0 => two real solutions]

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