# solve

• September 6th 2008, 02:11 AM
perash
solve
solve the equation

$x\sqrt {1 + \frac {1}{x}} + \frac {1}{x}\sqrt {x + 2} = 1 + \sqrt {x + \frac {2}{x} + 3}$
• September 6th 2008, 07:16 AM
Moo
Hello,

$
x\sqrt {1 + \frac {1}{x}} + \frac {1}{x}\sqrt {x + 2} = 1 + \sqrt {x + \frac {2}{x} + 3}
$

You can notice that $\left(1+\frac 1x\right) \cdot (x+2)=\left(x+\frac 2x+3\right)$

1. Good (?) method
$\text{equation}
\quad \Longleftrightarrow \quad \left(x \sqrt{1+\frac 1x}-1\right)-\left(\sqrt{\left(1+\frac 1x\right)(x+2)}-\frac 1x \sqrt{x+2}\right)=0$

$x \left(\sqrt{1+\frac 1x}-\frac 1x\right)-\sqrt{x+2} \left(\sqrt{1+\frac 1x}-\frac 1x\right)=0$

$(x-\sqrt{x+2})\left(\sqrt{1+\frac 1x}-\frac 1x\right)=0$

Therefore :

$\text{either } \left\{\begin{array}{ll} x-\sqrt{x+2}=0 \quad (1) \\ \sqrt{1+\frac 1x}-\frac 1x=0 \quad (2) \end{array} \right.$

$(1) \implies x=\sqrt{x+2} \implies x^2-x-2=0 \implies (x+1)(x-2)=0$
$\boxed{x=-1}$ or $\boxed{x=2}$

$(2) \implies \frac 1x=\sqrt{1+\frac 1x} \implies \frac 1{x^2}=1+\frac 1x \implies x^2+x-1=0$
$\boxed{x=\frac{-1 \pm \sqrt{5}}{2}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2. Rough method
Square both sides :

$x^2 \left(1+\frac 1x\right)+\frac{x+2}{x^2}+{\color{red}2 \sqrt{x+\frac 2x+3}}=1+\left(x+\frac 2x+3\right)+{\color{red}2 \sqrt{x+\frac 2x+3}}$

$(x^2+x)+\left(\frac 1x+\frac 2{x^2}\right)=1+x+\frac 2x+3$

$x^2-\frac 1x+\frac 2{x^2}-4=0$

Multiply by $x^2$ :

$x^4-4x^2-x+2=0$

Note that -1 is a solution, by looking at the coefficients.

$(x+1)(x^3-x^2-3x+2)=0$

Here, you can use the rational roots theorem... or see that 2 is a solution.

$(x+1)(x-2)(x^2+x-1)=0$
...
• September 6th 2008, 07:59 AM
flyingsquirrel
Quote:

Originally Posted by Moo
$(1) \implies x=\sqrt{x+2} \implies x^2-x-2=0 \implies (x+1)(x-2)=0$
$\boxed{x=-1}$ or $\boxed{x=2}$

$(2) \implies \frac 1x=\sqrt{1+\frac 1x} \implies \frac 1{x^2}=1+\frac 1x \implies x^2+x-1=0$
$\boxed{x=\frac{-1 \pm \sqrt{5}}{2}}$

In both cases $x$ can't be negative since we have $x=\sqrt{x+2}\geq0$ or $\frac 1x=\sqrt{1+\frac 1x}>0$.