This is Example 3 on page 12 of "Calculus: One and Several Variables" by Salas, Hille, Etgen (10th edition).

Solve the inequality

$$x^2 -2x + 5 \leq 0 $$

The example proceeds to complete the square to obtain the following result:

$$ x^2 -2x + 5 = ... = (x - 1)^2 + 4$$

I can follow just fine so far. The next line is what I have trouble with:

This tell us that

$$ x^2 -2x + 5 \geq 4 \qquad \text{ for all real } x, \tag{1}\label{1}$$

and thus there are no numbers that satisfy the inequality ... the solution set is the empty set $\emptyset$.

How does it tell us that? I don't understand how we can conclude the original equation is $\geq 4$.

I got the same solution, but by a slightly different way:

$$\begin{align*} x^2 - 2x + 5 &\leq 0 \\ x^2 - 2x + 1 - 1 + 5 &\leq 0 \\ (x - 1)^2 + 4 &\leq 0 \\ (x - 1)^2 &\leq - 4 \end{align*}$$

Since a square can't be less than $0$, I concluded the solution set is $\emptyset$.

But I still don't understand how \eqref{1} was concluded.

  • 3
    $\begingroup$ $(x-1)^2+4≤0\implies x^2-2x+5≤0$ so I don't understand the claim. $\endgroup$ – lulu May 18 at 20:30
  • $\begingroup$ As indicated by others, there must be a typo in the text. $\endgroup$ – imranfat May 18 at 20:32

Since x is real we know that $(x-1)^2\geq 0$. So, $(x-1)^2+4\geq4$ for all real $x$.

  • $\begingroup$ This explanation makes sense! Thanks! (But I still think the way it was written in the book was confusing...) $\endgroup$ – Calculemus May 18 at 21:14

Observe that

$$x^2-2x+5=(x-1)^2+4\ge4\implies x^2-2x+5\le0\;\text{ is never true}$$

The text is confuse...but one actually leads to the other.


What this is trying to say as currently written is that the inequality $$x^2-2x+5\le 0$$ has no real solutions.

This is because $$x^2-2x+5=(x-1)^2+4\ge 4$$ for all real $x$. And this is true because the real square $(x-1)^2\ge 0$.

The second part doesn't follow from the first, rather it follows from the fact that a real square is positive, and thus establishes a contradiction.


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