Hello,
Originally Posted by
Grandad
Note first that $\displaystyle AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1$
Let $\displaystyle x = \sqrt{A} + \sqrt{B}$
Then $\displaystyle x^2 = A + 2\sqrt{AB} + B = A +2+B$
... hence $\displaystyle x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},$
there is no need for the quadratic formula.
Originally Posted by
Grandad
$\displaystyle \Rightarrow (x^2)^2 =(A+B+2)^2= A^2 +2AB + B^2 + 4(A+B) + 4$
Noting that $\displaystyle A+B = x^2 - 2$, this simplifies to
$\displaystyle (x^2)^2 = 4x^2 +2\sqrt5 - 2$
I think the last equation is not correct. We have
$\displaystyle \begin{aligned}(x^2)^2&=(A+B+2)^2\\
&=(A+B)^2 + 4(A+B) + 4\\
&=(2\sqrt{5})^2+4(x^2-2)+4\\
&=4x^2+20-8+4\\
(x^2)^2&=4x^2+16\end{aligned}$
and the positive root of $\displaystyle X^2-4X^2-16$ is $\displaystyle \frac{4+\sqrt{16+4\times 16}}{2}=2+2\sqrt{5}$ so we get $\displaystyle x=+\sqrt{2+2\sqrt{5}}$ which matches the solution given above.