Prove that x²+x+1 is always positive.
The Correct Answer and Explanation is :
To prove that ( x^2 + x + 1 ) is always positive, we can use algebraic techniques, such as analyzing its discriminant and considering its behavior for all values of ( x ).
Step 1: Consider the function
We are looking at the quadratic function ( f(x) = x^2 + x + 1 ), which is a parabola opening upwards because the coefficient of ( x^2 ) is positive.
Step 2: Calculate the discriminant
The discriminant of a quadratic equation ( ax^2 + bx + c = 0 ) is given by ( \Delta = b^2 – 4ac ). For our equation, ( a = 1 ), ( b = 1 ), and ( c = 1 ), so the discriminant is:
[
\Delta = 1^2 – 4(1)(1) = 1 – 4 = -3
]
Since the discriminant is negative (( \Delta = -3 )), the quadratic equation has no real roots. This means that the function ( f(x) = x^2 + x + 1 ) does not cross the x-axis and does not take negative values for any real value of ( x ).
Step 3: Analyze the vertex
Since the quadratic opens upwards (because the leading coefficient is positive), the vertex of the parabola gives the minimum value of the function. The x-coordinate of the vertex is given by:
[
x = \frac{-b}{2a} = \frac{-1}{2(1)} = -\frac{1}{2}
]
Substitute ( x = -\frac{1}{2} ) into the function to find the minimum value:
[
f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} – \frac{1}{2} + 1 = \frac{3}{4}
]
Thus, the minimum value of ( f(x) = x^2 + x + 1 ) is ( \frac{3}{4} ), which is positive.
Step 4: Conclusion
Since the quadratic function ( f(x) = x^2 + x + 1 ) has no real roots, opens upwards, and has a minimum value of ( \frac{3}{4} ), it is always positive for all real values of ( x ). Therefore, we have proven that ( x^2 + x + 1 ) is always positive.