What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$?

Question

First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I

$\sqrt{x^2+1}- 2x+1>0$

$\sqrt{x^2+1}> 2x-1$

$0> 3x^2-4x$

$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.

I know that this solution is wrong, because I went and drew this graph. The correct result is $x\in (-\infty,\frac{4}{3}]$.

I'm totally confused on how to solve irrational inequalities now, because the official solving in my textbook looks like this:

$\sqrt{x^2+1}> 2x-1$. This inequality is fulfilled , if the right side is negative, therefore $x<\frac{1}{2}$. If $x\geq\frac{1}{2}$, the right side is positive or equals to $0$ and we get $0> 3x^2-4x$ which is true for $x\in (0,\frac{4}{3}]$. Now with previous condition $x\geq\frac{1}{2}$ we get the solution $[\frac{1}{2},\frac{4}{3})$.The complete solution set is $x\in (-\infty,\frac{4}{3}]$.

I never solved this type of inequality in this way, because it's messy- Why would I look at the conditions $x<\frac{1}{2}$ and $x\geq\frac{1}{2}$ when I can immediately tell where irrational part is defined and where not?

In the end, this textbook solution process only confused me. Could anyone please explain it why the correct solution is $(-\infty,\frac{4}{3}]$ or more concretely: Where did I miss the part of the solution $(-\infty,0]$?

Answer 1

You seem to believe that$$\sqrt{x^2+1}>2x-1\iff x^2+1>(2x-1)^2.\tag1$$This is false. The inequality $\sqrt{x^2+1}>2x-1$ holds automatically if $2x-1\leqslant0\left(\iff x\leqslant\frac12\right)$. Otherwise (that is, if $x>\frac12$) then, yes, the equivalence $(1)$ holds.

Answer 2

What you've overlooked is that it's not true that, assuming you know that $t \geq 0, \text{ then } \sqrt{t} \geq y \iff t \geq y^2$. What is true is that $\sqrt{t} \geq y \iff t \geq y^2 \lor y \lt 0$.

So you need to account for the possibility that $2x-1 \lt 0$.

Answer 3

When $A \geq 0,$ it is always true that $B > A$ is equivalent to $B^2 > A^2.$ But this is not always true when $A < 0.$

For example, if $A = -2$ and $B = -1,$ then $B > A$ is true but $B^2 > A^2$ is false.

The step where you squared both sides of $\sqrt{x^2+1}> 2x-1$ is only valid when $2x-1 \geq 0,$ that is, when $x \geq \frac12.$ For every other value of $x$ you have to use a different method.

One method that works for $x<\frac12$ is to notice that the right side of $\sqrt{x^2+1}> 2x-1$ is always negative while but the left side is never negative, therefore the left side is always greater than the right.

Answer 4

First of all, we should check for $x \ge \frac{1}{2}$ and $x < \frac{1}{2}$ because in this step: $$\sqrt{x^2+1}> 2x-1$$ we are squaring both sides. So sign of $2x-1$ may change the direction of inequality (precisely when $|2x-1| > \sqrt{x^2+1}$).

Now, for the case where $x < \frac{1}{2}$, we also have a solution for $$x^2+1 <4x^2-4x+1 \implies 3x^2-4x > 0 \implies x < 0 \lor x > \frac{4}{3}$$ but $x > \frac{4}{3}$ doesn't satisfy our first assumption $x < \frac{1}{2}$. Therefore, we get a solution $x < 0$ here.

Answer 5

If $A,B\in \Bbb R$,$$\sqrt A>B\iff \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A\ge 0\\ B\ge 0\\ A>B^2\end{cases}$$ In this instance $A=x^2+1$, $B=2x-1$ $$\begin{cases}x^2+1\ge 0\\ 2x-1<0\end{cases}\lor\begin{cases}x^2+1\ge 0\\ 2x-1\ge 0\\ x^2+1>4x^2-4x+1\end{cases}\iff x<\frac12\lor\begin{cases}x\ge \frac12\\ x(3x-4)<0\end{cases}\iff\\ \iff x<\frac12\lor \begin{cases}x\ge\frac12\\ 0<x<\frac43\end{cases}\iff x<\frac12\lor \frac12\le x<\frac43\iff x<\frac43$$

Answer 6

I'll try to explain in somewhat simpler terms than the other valid answers.

Also, I assume you really meant the strict inequality "$>$" and not "$\geq$".

The problem is in the squaring step, of course, and comes from the fact that the valid inequality $$1 > -1$$ does not imply the nonsensical squared inequality $$1^2 > (-1)^2.$$

Your task was to find all $x$ for which the original inequality holds, but you only found some because you missed the possibility that $\sqrt{x^2 + 1}$ is less positive than $(2x - 1)$ is negative (or equal), i.e. the case that $$ \sqrt{x^2 + 1} \leq -(2x - 1) . $$

The borderline case happens when $$ \sqrt{x^2 + 1} = |2x - 1|, $$ which has the two solutions you identified, $x_1 = 0$ and $x_2 = \frac{4}{3}$.

By inspection, $\sqrt{x^2 + 1}$ is less positive than $(2x - 1)$ is negative (or equal) for all $x \leq 0$.

Together with the part that you found, we have the solution set $$ (-\infty, \frac43). $$

If you meant "$\geq$" in the problem setup then the solution set is $(-\infty, \frac43]$.

Hope that makes sense.

Answer 7

The OP mentioned the solution given in the book, so here we express that logic.

$\quad \{ x \in \Bbb R \, | \, \sqrt{x^2+1} - 2x+ 1 \gt 0\} =$

$\quad\quad\quad \{ x \in (-\infty, \frac{1}{2}] \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \; \bigcup $

$\quad\quad\quad\{ x \in [\frac{1}{2},\frac{4}{3}] \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \; \;\;\;\;\,\bigcup$

$\quad\quad\quad \{ x \in [\frac{4}{3},+\infty) \; | \; \sqrt{x^2+1} - 2x+ 1 \gt 0\} \;=$

$\quad\quad\quad(-\infty, \frac{1}{2}] \; \bigcup \; [\frac{1}{2} , \frac{4}{3}) \; \bigcup \; \emptyset = (-\infty, \frac{4}{3}) $

Answer 8

You must consider: $$\sqrt{x^2+1}> 2x-1 \iff \\ 1) \ \sqrt{x^2+1}> 2x-1\color{red}{\ge 0} \ \ \text{or} \ \ 2) \ \sqrt{x^2+1}\ge \color{red}{0>} 2x-1$$ You square in $1)$, because both terms are nonnegative.

However, you must not square in $2)$, because you don't know how far away the LHS and RHS are from $0$ and you would impose an additional constraint. Moreover, once $0>2x-1$ holds, it automatically implies $\sqrt{x^2+1}>2x-1$. So: $$\sqrt{x^2+1}\ge \color{red}{0>} 2x-1 \iff \color{red}{0>}2x-1.$$

Hence, solution must be: $$x^2+1>(2x-1)^2 \ \ \text{or} \ \ 0>2x-1 \Rightarrow \\ x\in (0,\frac43] \ \ \text{or} \ \ x\in (-\infty,\frac12) \Rightarrow \\ x\in (-\infty,\frac43].$$