Why is the discriminant of quadratic f, Δ≤0, when f(x)≥0∀x∈R?

Question

Why is the discriminant of quadratic f, Δ≤0, when f(x)≥0∀x∈R?

Lindsey Rocha · Accepted Answer

Step 1Let&amp;nbsp;P(x)=ax2+bx+c&amp;nbsp;and suppose&amp;nbsp;P(x)≥0&amp;nbsp;for all x.Case 1:&amp;nbsp;a=0&amp;nbsp;In this case,&amp;nbsp;P(x)=bx+c&amp;nbsp;and&amp;nbsp;P(x)≥0&amp;nbsp;for all x, so it follows&amp;nbsp;b=0&amp;nbsp;(think of what the graph of a straight line looks like if&amp;nbsp;b≠0. So&amp;nbsp;b2−4ac=0Cae 2:&amp;nbsp;a&amp;gt;0&amp;nbsp;Then, we can complete the square,&amp;nbsp;P(x)=a(x+b2a)2+c−b24a&amp;nbsp;is&amp;nbsp;≥0&amp;nbsp;for all x, so in particular when&amp;nbsp;x=−b2a, and thus&amp;nbsp;c−b24a≥0Rearranging yields&amp;nbsp;b2−4ac≤0&amp;nbsp;(where did I use the fact&amp;nbsp;a&amp;gt;0?)This completes the proof (Why is case 3 of&amp;nbsp;a&amp;lt;0&amp;nbsp;not possible

Nigel Nichols · Accepted Answer

Step 1If D&amp;lt;0, this means that the quadratic has no real roots. This forces us to conclude that the graph of the function must either lie completely above, or completely below the x-axis, since it can neither touch nor intersect it. Since the coefficient of x2 is positive, this means that the parabola must be upward-opening (because any curve of this nature is of the form a(x+b)2+c, hence can be formed by shifting and compressing/expanding the graph of y=x2). So it can&#039;t lie below the x-axis, hence it must lie above.

Why is the discriminant of quadratic \(\displaystyle{f},\

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