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I want to ask about radical inequality problem.

Here's the question:

Find the solution sets for $\sqrt{x^2+12x+35}\geq x-10$

My attempts to tackle this problem is like this:

Firstly I try to squaring this inequality such that:

$x^2+12x+35\geq (x-10)^2$

$x^2+12x+35\geq x^2-20x+100$

$x=\frac{65}{32}$

And then I apply the condition for the form under the radical, such that:

$x^2+12x+35\geq 0$ and also for the right hand, such that: $x-10\geq 0$

I solve for both of them and got $x\leq -7$ and $x\geq -5$ for the first one and $x>=10$ for the right hand.

Then, I combine all of the solutions to get whole solutions $x>=10$

From my number line, I can conclude that the solution should be $x\geq 10$

But, when I'm trying to check it in wolfram, I got the solution should be $x\leq 7$ or $x\geq -5$

Please somebody explain to me about this difference, so I can get the right answer from this problem. Thanks anyway.

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    $\begingroup$ Why the condition $x-10\ge0$? If $x-10<0$ and the radical is well defined, surely the inequality holds, right? $\endgroup$
    – Did
    Sep 15, 2017 at 10:49
  • $\begingroup$ First you have to sate the constraint on your inequality . This inequality make senses only when term under square root is nonnegative . Add9ng to this it becomes obvious when the left hand side is non-positif. $\endgroup$
    – Guy Fsone
    Sep 15, 2017 at 10:52
  • $\begingroup$ @Did, I think the radical inequallty must have a condition for right hand side $a>= 0$ , such that this inequality can be holds? Maybe, you can explain to me about your statement? thanks $\endgroup$
    – akusaja
    Sep 15, 2017 at 10:57
  • $\begingroup$ You have 1200 rep and haven't learned $\geq$ yet? Wow... just wow. $\endgroup$ Sep 15, 2017 at 11:07
  • $\begingroup$ akusaja What? Sorry but did you even read my comment? I mean, carefully. $\endgroup$
    – Did
    Sep 15, 2017 at 11:17

4 Answers 4

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First you have to sate the constraint on your inequality . This inequality make senses only when term under square root is nonnegative . i.e $$x^2+12x+35\ge 0\Longleftrightarrow (x+5)(x+7)\ge 0\Longleftrightarrow x\in(-\infty, -7]\cup [-5, \infty)$$

Adding to this it becomes obvious when the left hand side is non-positif. ie if

$$ x-10\le 0 \Longleftrightarrow x\in (-\infty ,10]$$

So, if $x\in (-\infty ,10]\cap((-\infty, -7]\cup [-5, \infty)) =(-\infty, -7]\cup [-5,10]$ we have $\sqrt{x^2+12x+35} >= x-10$

Now if $x\in (-\infty, -7]\cup [-5, \infty) $ and $x\in [10, \infty)$ which means $x\in [10, \infty) $ and $x-10\ge0$ you can proceed as you did i.e

For $x\in [10, \infty)$ we have,

$$\sqrt{x^2+12x+35} >= x-10 \Longleftrightarrow x^2+12x+35\ge x^2-20x+100\Longleftrightarrow 32x \ge 65$$ Then, $x\ge 65/32 =2.03125$ and $x\in [10, \infty)$ this implies that $x\in [10, \infty)$

finally, $\sqrt{x^2+12x+35} >= x-10 \Longleftrightarrow x\in [10, \infty)\cup (-\infty, -7]\cup [-5,10] =(-\infty, -7]\cup [-5,\infty)$

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  • $\begingroup$ These numbers should be $-5$ and $-7$ instead of $5$ and $7$. $\endgroup$
    – B. Goddard
    Sep 15, 2017 at 11:17
  • $\begingroup$ you are right I fixed it now $\endgroup$
    – Guy Fsone
    Sep 15, 2017 at 11:38
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We can write your equations as $$\sqrt{(x+6)^2-1}\ge x-10$$

For us to be able to write $$\sqrt{(x+6)^2-1}$$

It must be that $$ \bbox[5px,border:2px solid red]{x\in \Big(-\infty ,-7\Big]\cup \Big[-5,\infty \Big)}$$ $$$$$$$$ For $$ -5\le x\le 10$$ $$\sqrt{(x+6)^2-1}\ge 0\ge x-10$$ Which is true.

Hence, $$ \bbox[5px,border:2px solid red]{x\in \Big[-5,10\Big]}$$ $$$$$$$$ For $x\gt 10$ $$\sqrt{(x+6)^2-1}\ge x-10$$ $$(x+6)^2-1\ge (x-10)^2$$ $$(x+6)^2-(x-10)^2 \ge 1$$ $$((x+6)+(x-10))((x+6)-(x-10))\ge 1$$ $$(2x-4)16\ge 1$$ $$32x-64\ge 1$$ $$32x\ge 65$$ $$ x\ge \frac{65}{32}$$ But $$ \bbox[5px,border:2px solid red]{x\in \Big[10,\infty \Big)}$$

Considering all conditions, $$x\in \Big(-\infty,-7\Big]\cup \Big[-5,\infty\Big)$$

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Hints:

  • First, you have to determine the domain of validity of the inequation, i.e. for which values of $x$ the radicand is non-negative. As $$x^2+12x+35=(x+5)(x+7),$$ this domain is $\;(-\infty,-7]\cup[-5,+\infty)$.

  • Second, you must remember that, on the domain of validity, $$\sqrt A\ge B\iff A\ge B^2\quad\textbf{or}\quad B\le 0.$$

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  • $\begingroup$ Upvoting the only reasonably sane answer in this circus. $\endgroup$
    – B. Goddard
    Sep 15, 2017 at 11:26
  • $\begingroup$ @bernard, so the condition for the right hand side is should be less than zero? not larger than zero? thanks $\endgroup$
    – akusaja
    Sep 15, 2017 at 11:32
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    $\begingroup$ It is one possibility, provided the l.h.s. is defined. $\endgroup$
    – Bernard
    Sep 15, 2017 at 11:39
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Let $x^2+12x+35=y^2$, where $y\geq0$.

Thus, $x=-6-\sqrt{y^2+1}$ or $x=-6+\sqrt{y^2+1}$ and we need to solve $$y+16\geq\sqrt{y^2+1}$$ and we need to solve

$$y+16\geq-\sqrt{y^2+1}.$$ But both inequalities are obviously true for all $y\geq0$,

which says that the starting inequality is equivalent to $$x^2+12x+35\geq0,$$ which gives the answer: $$(-\infty,-7]\cup[-5,+\infty).$$ Done!

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