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*}
How do we solve the equation?
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algebra-precalculus
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1$x=2$ seems to satisfy the equation and i don't see any other way other than squaring "carefully" getting six-degree equation – 2012-09-26