Question

If $m$ and $n$ are natural numbers such that $\sqrt{7+\sqrt{48}}=\sqrt{m}+\sqrt{n}$, then $m^2+n^2$ equals

• A. $25$
• B. $37$
• C. $29$
• D. $41$

Answer 1

The correct option is A $25$

Given $\sqrt{7+\sqrt{48}}=\sqrt{m}+\sqrt{n}$.
Square both side,
we will get
$7+\sqrt{48}=(\sqrt{m}+\sqrt{n}{)}^2$ [$(a+b{)}^2=a^2+b^2+2(a\times b)$ ]
$7+\sqrt{48}=(m+n+2\sqrt{m\times n})$
now comparing L. H. S and R. H. S
$m+n=7$ .......... equ($1$)
$m\times n=12$ .......equ($2$) [because $\sqrt{48}=2\times \sqrt{12}$]
now, from equ($1$) $n=7-m$ ........ equ($3$)
on solving equ($2$) and equ($3$), we will get
$m^2-7m+12=0$
on solving this equation we will get
$m=4or3$
putting the value of $m$ in equ($3$)
n=$3or4$
hence $m^2+n^2=4^2+3^2=25$