How do we solve the equation?

Question

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*}

Answer 1

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.. $$(a^2-1)a+(a^2+4)\cdot\sqrt{a^2+5} = (a^2+1)(a^2+3)$$

Answer 2

try squaring everything and expanding then reducing ...

Answer 3

Put \begin{equation*} f(x) = \dfrac{x+1}{\sqrt{x+2} + 2} + \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4. \end{equation*} First case. $x \geqslant -1$. We have \begin{align*} \dfrac{x+1}{\sqrt{x+2} + 2} + \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4 &< \dfrac{x+1}{ 2} + \dfrac{x+6}{ 3} - x - 4 \\ &= \dfrac{-x - 9}{6} < 0. \end{align*} Second case. $-2 \leqslant x \leqslant - 1$, then $\dfrac{x + 1}{\sqrt{x + 2} + 2} \leqslant 0$.

Put $$g(x) = \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4.$$ We have \begin{equation*} g'(x) = \dfrac{1}{\sqrt{x + 7} + 3} - 1 - \dfrac{x + 6}{2\sqrt{x + 7}(\sqrt{x + 2} + 3)^2} < 0, \quad \forall -2 \leqslant x \leqslant - 1. \end{equation*} Therefore $g$ is a decreasing function. Then $g(x) \leqslant g(-2) = 1 - \sqrt{5} < 0$. From the two cases, we get 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 $x = 2.$