On what natural values of a, the equation has no roots

Question

$$\left|x+9\right|=\frac{x}{2}+a$$ $$$$ $$ \left(x+9\right)=\frac{x}{2}+a,\quad x=2a-18\tag{1}$$ $$ -\left(x+9\right)=\frac{x}{2}+a,\quad x=-\frac{2a+18}{3}\tag{2}$$ And i don't know what to do next?

Answer 1

For $$x\geq -9$$ we get

$$x+9=x/2+a$$ so

$$x=2(a-9)$$ and we get

$$2(a-9)\geq -9$$ so we get

$$a\geq \frac{9}{2}$$ Can you finish? So no Solutions for $$a<\frac{9}{2}$$

The other case:

$x<-9$ then we get

$$x=\frac {2(-9-a)}{3}$$ and it must be

$$\frac{2(-9-a)}{3}<-9$$ this gives

$$-18-2a<-27$$ thus $$\frac{9}{2}<a$$

therefore we get no Solutions for $$\frac{9}{2}\geq a$$

Answer 2

Discuss with the value of $x$.

(1) when $x+9\geq0$, then you have already got $x=2a-18$. By the assumption, we have $$ x+9\geq0\Rightarrow 2a-18+9\geq0\Rightarrow a\geq9/2 $$ (2) when $x+9<0$, then $x=-(2a+18)/3$. Thus, $$ x+9<0\Rightarrow -\frac{2}{3}a-6+9<0\Rightarrow a>18 $$ Consequently, when $a<\frac{9}{2}$, there dose not exist $x$ such that the equation holds.