If $2 |3-x| + 5 = k$ has 2 $\mathbb{R}$ solutions for $x$, what are the possible values for $k$?

Question

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 $k$?

Answer 1

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

Answer 2

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, $\displaystyle k>5$

Answer 3

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

Answer 4

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...