# solve

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

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

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

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

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

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

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

Therefore :

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

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

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

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2. Rough method
Square both sides :

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

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

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

Multiply by $\displaystyle x^2$ :

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

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

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

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

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

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

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

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