Determine limit of |a+b| This is a simple problem I am having a bit of trouble with. I am not sure where this leads.

Given that $\vec a = \begin{pmatrix}4\\-3\end{pmatrix}$ and $|\vec b|$ = 3, determine the limits between which $|\vec a + \vec b|$ must lie.

Let, $\vec b = \begin{pmatrix}\lambda\\\mu\end{pmatrix}$, such that $\lambda^2 + \mu^2 = 9$
Then, 
$$
\begin{align}
\vec a + \vec b &= \begin{pmatrix}4+\lambda\\-3 + \mu\end{pmatrix}\\
|\vec a + \vec b| &= \sqrt{(4+\lambda)^2 + (\mu - 3)^2}\\
&= \sqrt{\lambda^2 + \mu^2 + 8\lambda - 6\mu + 25}\\
&= \sqrt{8\lambda - 6\mu + 34}
\end{align}
$$
Then I assumed $8\lambda - 6\mu + 34 \ge 0$. This is as far I have gotten. I tried solving the inequality, but it doesn't have any real roots? Can you guys give me a hint? Thanks.

We know that $\|a\|=5$, $\|b\|=3$, and we have two vector formulas
$$ \|a+b\|^2=\|a\|^2+2(a\cdot b)+\|b\|^2,$$
$$ a\cdot b = \|a\| \|b\| \cos\theta.$$
Combining all this, we have
$$\|a+b\|^2 = (5^2+3^2)+2(5)(3)\cos\theta.$$
Cosine's maximum and minimum values are $+1$ and$-1$, so we have 
$$\|a+b\|^2 \in [4,64]$$
$$\|a+b\| \in [2,8].$$
