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So we have $2 |3-x| + 5 = k$, where $k$ is a constant. Provided this equation has two real solutions for $x$, what is the range of possible values for $k$?

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up vote 2 down vote accepted

Your equation is equivalent to $|3-x|=\frac{k-5}2$ which will have two real roots if $\frac{k-5}2>0\iff k>5$.

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Why does $\frac{k-5}2$ have to be $> 0$ for $x$ to have two real solutions? Also, my textbook has $k \leq 11$ as a further condition. – hb20007 May 21 '14 at 16:42
$|y|=a\iff y=a\vee y=-a$, so it will have $2$ real roots iff $a>0$. – Jlamprong May 21 '14 at 16:45
The other answers were longer but I'm picking this as the best answer because you explained the logic behind it in your comment. Thanks – hb20007 May 21 '14 at 16:58
You are very wellcome. If you accept my answer, please check it... :D – Jlamprong May 21 '14 at 17:00
but first I think there was a mistake in the term in the modulus... I sent you an edit – hb20007 May 21 '14 at 17:02

$2|3-x|+5$ has a minimum at $x=3$ and is $\text V$ shaped above it. Hence, $k>5.$

enter image description here

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very nicely explained. – SA-255525 May 21 '14 at 17:11

We have $\displaystyle\frac{k-5}2=|3-x|\ge0\iff k-5\ge0\iff k\ge5$

If $\displaystyle k=5,|x-3|=0\iff x=3$ for real $x$

So, $\displaystyle k>5$

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oops I just edited the question to correct a mistake about what's in the modulus sign – hb20007 May 21 '14 at 16:37
@hb2 Same line of reasoning holds though... – gebruiker May 21 '14 at 16:37
@hb20007, For real $x,|x|=+x$ if $x\ge0$ and $|x|=-x$ for $x<0$ – lab bhattacharjee May 21 '14 at 16:38
@labbhattacharjee My textbook also has $k \leq 11$ as a further condition? – hb20007 May 21 '14 at 16:44
@hb20007, Counter example, $k=13\implies |3-x|=4\implies 3-x=\pm4\implies x=?$ – lab bhattacharjee May 21 '14 at 16:47

We have $ \displaystyle\frac{k-5}2=|3-x|\ge0\Rightarrow k-5\ge0\Rightarrow k\ge5$

  1. If $\displaystyle k=5 $ we have $|x-3|=0\Rightarrow x=3$ for real $x$
  2. If $k>5$ we have $\displaystyle|x-3|=\frac{k-5}2\Rightarrow-\frac{k-5}2-3=x\vee x=\frac{k-5}2+3$

Since you requested we have two (real) roots, we must have $\displaystyle k>5$.

NB. Your edit that we should have $k\leq11$ somehow is not something that follows from the equation you've given us...

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Thank you for the great answer. I have already chosen a best answer so +1 is all I can give – hb20007 May 21 '14 at 17:04
@hb2 Not that I'm trying to score some easy reputation points, but I believe that you can alter that decision whenever you like... – gebruiker May 21 '14 at 17:07

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