How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$?

Question

Could someone explain how to get from: $x-\frac{1}{x}=A$ to $x+\frac{1}{x}=\sqrt{A^2+4}$ ? It is one of the Algebra II tricks.

Thanks.

Answer 1

Start by squaring both sides: $$\begin{align}x-\frac{1}{x}&=A\\\left(x-\frac{1}{x}\right)^2&=A^2\\x^2-2+\frac{1}{x^2}&=A^2.\end{align}$$ Then try adding $4$ to both sides and "reversing" the processes above.

Answer 2

Ok, I got it!

$x-\frac{1}{x}=A$

$\Bigl(x-\frac{1}{x}\Bigl)^2=A^2$

$x^2+\frac{1}{x^2}-2=A^2$

$x^2+\frac{1}{x^2}=A^2+2$

$\Bigl(x+\frac{1}{x}\Bigl)^2-2=A^2+2$

$\Bigl(x+\frac{1}{x}\Bigl)^2=A^2+4$

$\sqrt{\Bigl(x+\frac{1}{x}\Bigl)^2}=\sqrt{A^2+4}$

$x+\frac{1}{x}=\sqrt{A^2+4}$

Answer 3

Remark that $x-\frac 1x=A\iff x^2-Ax-1=0$

Which is a quadratics with roots $x$ and $-\frac 1x$.

Calculating $\Delta=A^2+4$ and roots $\dfrac{A\pm\sqrt{\Delta}}{2}$

It is now obvious why their sum is $A$ and their difference is $\sqrt{\Delta}$.