how can i put this equation in function of $B$: $A + \sqrt{A^2 + const} = B$

Question

$A + \sqrt{A^2 + const} = B$

Can somebody put this equation in function of $B$. Something like: $A = B \times something$

thank you all :D

Answer 1

HINT: rewriting the equation as $$\sqrt{A^2+C}=B-A$$ after squaring we have $$A^2+C=B^2+A^2-2BA$$ therefore $$A=\frac{B^2-C}{2B}$$

Answer 2

HINT:

$$\implies B-A=\sqrt{A^2+\text{constant}}$$

Squaring both sides

$$(B-A)^2=A^2+\text{constant}$$

Can you take it from here?

Answer 3

In complete case you can follow this:

$$(B-A)^2 = A^2 + c\Rightarrow B^2 + A^2 - 2AB = A^2 + c \Rightarrow 2AB = B^2- c \Rightarrow A = \frac{B}{2} -\frac{c}{2B} = B(\frac{1}{2} - \frac{c}{2B^2}) = B \times something$$