Integral values of an expression

Question

Let $b=\sqrt{a^2+5a+8}-\sqrt{a^2-3a+4}$

Find number of integral values of b.

My $long$ way using Calculus :

Find domain of function : $R$

Note that function is continuous

Prove the function is always increasing

Find limit as $x -> \infty$ and $x->-\infty$

Get it as $-4$ and $4$

Conclude that only possible integral values are ${-3,-2,-1,0,1,2,3}$.

But I am not happy as that is too long for a 3 marker question (in my exam)

I want nice way perhaps using theory of quadratic expressions?

EDIT Well you guys gave a short proof using calculus but I wanted to see a proof without using calculus as it was not in the syllabus of the test. And I have definitely not studied about asymptotes in detail (only in hyperbola briefly)

Both of the answers are excellent.

Answer 1

The asymptotics can be seen by completing the square:

$$a^2+5a+8=\left(a+\frac 52\right)^2+\frac 74$$

$$a^2-3a+4=\left(a-\frac 32\right)^2+\frac 74$$

So for large positive $a$ the expression is close to $\left(a+\frac 52\right)-\left(a-\frac 32\right)=4$, and for large negative $a$ it is close to $-\left(a+\frac 52\right)+\left(a-\frac 32\right)=-4$

Then you are done if you can show the function is continuous and increasing.

Answer 2

Just note the discriminants $5^2-4\cdot8=-7$ and $(-3)^2-4\cdot4=-7$ are negative, so the quadratics are always positive, therefore the function is continuous and you can write it like this: $$\sqrt{a^2+5a+8}-\sqrt{a^2-3a+4}=\frac{8a+4}{\sqrt{a^2+5a+8}+\sqrt{a^2-3a+4}}$$ Now you see the limits are $\frac{-8}{1+1}=-4$ for $a\to-\infty$ and $\frac8{1+1}=4$ for $a\to+\infty$, so the integral values are $-3, -2, -1, 0, 1, 2, 3$ and you have to exclude $-4$ and $4$:

$$\begin{align}\sqrt{a^2+5a+8}+\sqrt{a^2-3a+4}&=\sqrt{\left(a+\frac52\right)^2+\frac74}+\sqrt{\left(a-\frac32\right)^2+\frac74}\\&>{\left|a+\frac52\right|+\left|a-\frac32\right|}\geq|2a+1|.\end{align}$$ This means $\left|\frac{8a+4}{\sqrt{a^2+5a+8}+\sqrt{a^2-3a+4}}\right|<4$ and you don't need to prove the function is increasing.

Another Way:

Substitute $2a=b-1$: $${\sqrt{a^2+5a+8}-\sqrt{a^2-3a+4}=\frac12\left(\sqrt{(2a)^2+10\cdot2a+32}-\sqrt{(2a)^2-6\cdot2a+16}\right)}\\\begin{align}&=\frac12\left(\sqrt{(b-1)^2+10(b-1)+32}-\sqrt{(b-1)^2-6(b-1)+16}\right)\\&=\frac12\left({\sqrt{b^2+8b+23}-\sqrt{b^2-8b+23}}\right)\end{align}$$ Similarly like in the first solution $$\begin{align}\left|\frac12\left({\sqrt{b^2+8b+23}-\sqrt{b^2-8b+23}}\right)\right|&=\frac12\frac{|16b|}{\sqrt{(b+4)^2+7}+\sqrt{(b-4)^2+7}}\\&<\frac{|8b|}{|b+4|+|b-4|}\leq\frac{|8b|}{|2b|}=4\end{align}$$ This also proves the square roots are well defined.

So without loss of generality $b\geq0$ and for $n\in\mathbb N$ assuming $0\leq n\leq3$, we want to solve $$\frac12\left({\sqrt{b^2+8b+23}-\sqrt{b^2-8b+23}}\right)=n\\\begin{align}&\Longleftrightarrow 2(b^2+23)-2\sqrt{(b^2+23+8b)(b^2+23-8b)}=4n^2\\&\Longleftrightarrow b^2+23-2n^2=\sqrt{(b^2+23)^2-64b^2}\\&\Longleftrightarrow(b^2+23-2n^2)^2=(b^2+23)^2-64b^2\\&\Longleftrightarrow-4n^2(b^2+23)+4n^4=-64b^2\\&\Longleftrightarrow(16-n^2)b^2=23n^2-n^4\\&\Longleftrightarrow b^2=\frac{n^2(23-n^2)}{16-n^2}\end{align}$$ The 3rd equivalence holds because $b^2+23-2n^2\geq b^2+5>0$ and the last holds because $n^2<16$, so $-3, -2, -1, 0, 1, 2, 3$ are all integral values.