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Prologue: $a > b\implies a^2>b^2$ will hold only when $|a| > | b|$.

So I was messing with this inequality for quite some time

$$\sqrt{8+2x-x^2}>6-3x$$

  • First things first I found out that the common domain of this inequality is actually $[-2,4]$.

  • when $6-3x <0$ or $x>2$ the inequality holds for $(2,\infty)$

When $6-3x \leq0$, the author squared the inequality or in other words " $a^2>b^2$ followed" which in turn implied that

$$\left|\sqrt{ 8+2x-x^2}\right| > |6-3x|$$ holds in order for "$>$" to remain intact.

How do I be sure about this?

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    $\begingroup$ I think you are stuck at why when $6-3x \geq 0$ (rather than $<$ which you wrote), the author could square the inequality on both sides. This is because if both $a$ and $b$ are $\geq 0$, then $$a^2>b^2\implies a>b$$, hence if you know $6-3x\geq 0$, you can square both sides and solve $$8+2x-x^2\geq (6-3x)^2$$ to get your answer, provided the left hand side is in the correct domain $\endgroup$ – Lazy Lee Jul 25 '17 at 14:54
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    $\begingroup$ $a>b\implies a^2 > b^2$ is also true when $a\le b$, just saying. $\endgroup$ – peterwhy Jul 25 '17 at 15:06
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$$\sqrt{8+2x-x^2}>6-3x\tag{*}$$

  • First things first I found out that the common domain of this inequality is actually $[-2,4]$.

  • when $6-3x <0$ or $x>2$ the inequality holds for $(2,\infty)$

Note that the second bullet point is wrong. As you have observed, $$ 8+2x-x^2\geq 0 $$ if and only if $x\in[-2,4]$. This means that the solution set to ($*$) must be a subset of $S:=[-2,4]$ and one should not bother with any point outside this set. Now there are obviously two cases to consider:

  • $x\in[-2,4]$ and $6-3x<0$; namely, $x\in(2,4]$.
  • $x\in[-2,4]$ and $6-3x\geq 0$; namely, $x\in[-2,2]$.

For the first case, one can see that any $x\in(2,4]$ is a solution to ($*$).

For the second case, note that you are comparing two nonnegative real numbers and for $a,b\geq 0$,

$a>b$ if and only if $a^2>b^2$.

Thus when $x\in[-2,2]$, ($*$) is equivalent to $$ 8+2x-x^2>(6-3x)^2. $$

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  • $\begingroup$ Listen when x >2 , let x=3, * gives root 5 > -3 which is true. $\endgroup$ – user33699 Jul 25 '17 at 15:34
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    $\begingroup$ So, what is your point? $\endgroup$ – Jack Jul 25 '17 at 15:35
  • $\begingroup$ Imagine graphically. The term with squareroot lies entirely above x axis.Also as 6-3x is a straight line with negetive slope, with x axis intercept at x=2. Beyond that it gives negetive output for entire set(2,infinity).so this is also a "part" of solution. $\endgroup$ – user33699 Jul 25 '17 at 15:39
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    $\begingroup$ I have no idea what you are talking about. Any point outside the interval $[-2,4]$ can not be a solution because $8+2x-x^2<0$, which makes the expression $\sqrt{8+2x-x^2}$ nonsense. $\endgroup$ – Jack Jul 25 '17 at 15:40
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    $\begingroup$ You are welcome. $\endgroup$ – Jack Jul 25 '17 at 15:59

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