Please could someone advise me on solving this:
If √m +√n = √(7+√48) calculate to the nearest digit m^{2} + n^{2 ? Thanks }
Hello, ACM!
$\displaystyle \text{If }\sqrt{m} + \sqrt{n} \:=\:\sqrt{7+\sqrt{48}},\,\text{calculate to the nearest digit }m^2+n^2.$
Whenever I see that an expression under a square root contains a square root,
. . I always wonder if that expression is a square.
Inside, we have: .$\displaystyle 7 + \sqrt{48} \:=\:7 + 4\sqrt{3}$ . . . . which happens to be $\displaystyle (2 + \sqrt{3})^2$
The problem becomes: .$\displaystyle \sqrt{m}+\sqrt{n} \;=\;\sqrt{(2 + \sqrt{3})^2} $
. . . . . . . . . . . . . . . . . .$\displaystyle \sqrt{m}+\sqrt{n} \;=\;2 + \sqrt{3}$
Hence: .$\displaystyle \begin{Bmatrix}m &=& 4 \\ n &=& 3\end{Bmatrix}$ . or vice versa.
Therefore: .$\displaystyle m^2 + n^2 \;=\;4^2 + 3^2 \;=\;25$