Question

Suppose that, ${\log }_{10}(x-2)+{\log }_{10}y=0$ and $\sqrt{x}+\sqrt{y-2}=\sqrt{x+y}$.Theh the value of $x+y$, is:

• A. $2$
• B. $2\sqrt{2}$
• C. $2+2\sqrt{2}$
• D. $4+2\sqrt{2}$

Answer 1

Answer: (C)

$2+2\sqrt{2}$

${\log }_{10}x-2+{\log }_{10}y=0$

$\Rightarrow {\log }_{10}(x-2)y=0$

$\Rightarrow (x-2)y=1$

$\Rightarrow xy=1+2y$ (equation $1$)

$\sqrt{x}+\sqrt{y-2}=\sqrt{x+y}$

Squaring both side,

$\Rightarrow x+(y-2)+2\sqrt{x(y-2)}=(x+y)$

$\Rightarrow 2\sqrt{x(y-2)}=2$

$\Rightarrow \sqrt{x(y-2)}=1$

$\Rightarrow x(y-2)=1$

$\Rightarrow xy=(1+2x)$ (equation $2$)

From equation $1$ and $2$

$(x=y)$

Putting this in equation $1$

$x^2=1+2x$

$(x^2-2x-1)=0$

$x=1+\sqrt{2},1-\sqrt{2}$

$y=1+\sqrt{2},1-\sqrt{2}$

But $x,y$ both should be greater than 2.

$\therefore (x+y)=2+2\sqrt{2}$