Inequalities with no solution/real number solution

Question

I have this inequality: $x^2 -4x +400>0.$

I realize that I cannot solve for $x$, however, when I plug the inequality into desmos, the solution seems to be all real numbers. But how do I show this/prove this mathematically?

Answer 1

There are several ways that come to mind.

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First Way:

• Solve the equation $x^2 - 4x + 400 = 0$, i.e. the equality case.
• In this case, you conclude there are no real roots.
• Hence, the parabola can never cross the $x$-axis, i.e. it lies entirely above or entirely below the axis.
• Plug in an $x$-value you like, say, $x=0$. If the result is $>0$, then the parabola is above the $x$-axis, and if the result is $<0$, then the parabola is below.
• If above the axis, then each real number $x$ is a solution; if below the axis, no real number $x$ is a solution.

(This somewhat implicitly uses a result from calculus known as the intermediate value theorem, but the ideas can be deduced for parabolas quite simply.)

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Second Way:

• Note that $$x^2 - 4x + 400 = x^2 - 4x + 4 + 396 = (x-2)^2 + 396$$
• This is the sum of two numbers, one nonnegative (since $z^2 \ge 0$) and one strictly positive.
• What is their sum always going to be? This result is independent of the value of $x$ chosen.

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Third Way:

• Find the vertex of the parabola. If given a parabola $f(x) = ax^2 + bx + c$, then the vertex is at $$ (x,y) = \left( \frac{-b}{2a}, f \left( \frac{-b}{2a} \right) \right) $$
• If the parabola opens up, the vertex is the absolute minimum of the graph, and the parabola never gets smaller than the $y$-coordinate. Likewise, if the parabola opens down, the vertex is its absolute maximum, and you will never get larger than the $y$-coordinate.
• One can deduce that your parabola opens up since the coefficient of $x^2$ is positive. Find the $y$-coordinate of your vertex, and therefore conclude that the parabola takes on exclusively positive values (since the $y$-coordinate of the vertex is positive, and all values must exceed this).

(One can justify this with calculus as well, instead of merely thinking of the geometric properties of a parabola: the derivative is zero at the vertex, positive on one side, and negative on the other, leading to results about extrema via the first derivative test. Or one can use the second derivative test, noting that the parabola is concave-up or concave-down.)