# How do we solve the equation?

How do we solve the following equation in the set of real numbers? $$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$ I wrote the given equation has the form \begin{equation*} (x+1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4) \end{equation*} This equation is equivalent to \begin{equation*} (x-2)\left(\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4\right) = 0. \end{equation*} But I can not prove that the equation \begin{equation*} \dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 = 0 \end{equation*} has no solution. Detail \begin{equation*} \dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 <0, \forall x \geqslant -2. \end{equation*}

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$x=2$ seems to satisfy the equation and i don't see any other way other than squaring "carefully" getting six-degree equation –  Aang Sep 26 '12 at 7:31

Not much simpler, but you can also try something like $a:=\sqrt{x+2}$ then $\sqrt{x+7}=\sqrt{a^2+5}$, then it will have only one sqrt in the equation, put that on one side and the rest on the other side.. $$(a^2-1)a+(a^2+4)\cdot\sqrt{a^2+5} = (a^2+1)(a^2+3)$$

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That's a good technique! I'll definitely remember it. +1 –  Daniel Littlewood Sep 26 '12 at 15:19
Defining $y=x+1$ simplifies things a little $\ldots$ –  bwsullivan Sep 26 '12 at 7:56
We have $$\dfrac{x+1}{\sqrt{x+2}+2}<\dfrac{x+2}{\sqrt{x+2}+2}\leqslant \dfrac{x+2}{2}$$ and $$\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+6}{3}\leqslant \dfrac{x+6}{2}.$$ Therefore $$\dfrac{x+1}{\sqrt{x+2}+2}+\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+2}{2}+\dfrac{x+6}{2}=x+4$$ Thus the equation \begin{equation*} \dfrac{x+1}{\sqrt{x+2} + 2} + \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4 = 0 \end{equation*} has no solution. Thus, the given equation has only root $x = 2.$