Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$

Question

I have to simplify the following expression:

$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$

Answer: $\sqrt{a+b}-\sqrt{a-b}$

I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)

Answer 1

Because $$\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}=\frac{\left(\sqrt{a+b}\right)^2-\sqrt{(a-b)(a+b)}}{\sqrt{a+b}}=\sqrt{a+b}-\sqrt{a-b}$$

The domain is $a+b>0$ and $(a-b)(a+b)\geq0,$ which gives $a+b>0$ and $a-b\geq0.$

Answer 2

You can write $$\frac{(a+b)^2-(a^2-b^2)}{\sqrt{a+b}(a+b+\sqrt{a^2+b^2})}$$

Answer 3

The constraints are indeed $$\begin{cases}a^2-b^2\ge0,\\a+b>0\end{cases}$$

which is equivalent to

$$\begin{cases}a-b\ge0,\\a+b>0\end{cases}$$ or

$$\begin{cases}a\ge b,\\a>-b.\end{cases}$$

Unfortunately, you cannot summarize with $a>|b|$ nor $a\ge|b|$ because of the asymmetry.

Answer 4

1) You have $\sqrt{a + b}$ in the denominator before simplification, so you can not have $\sqrt{a+ b} = 0$. So $a+b \ne 0$ and $a \ne -b$.

2) You have $\sqrt{a+b}$ before simplification so $a+b \ge 0$ and by 1) you have $a+b \ne 0$ so $a+ b > 0$ or $a > -b$.

3) You have $\sqrt{a^2 - b^2}$ before simplification so $a^2 -b^2 \ge 0$.

$a^2 - b^2 = (a+b)(a-b)$ and by 2) we have $a+b > 0$ so $a-b \ge 0$ so $a \ge b$.

We have $a \ge b$ and $a > -b$ so $a \ge |b|$.

If $b < 0$ then $-b > 0$ and $a > -b > 0$. If $b=0$ then $a > -b = 0$. If $b> 0$ then $a \ge |b| =b > 0$ so we have:

$a > 0$ and $a \ge |b|$ and if $b < 0$ then $a > |b|$.

...

Alternatively $a^2 -b^2 \ge 0$ so $a^2 \ge b^2$ so $|a| \ge |b|$.

but $a + b > 0$ so $a > -b$ and .... same results of above.