Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$

Find all real roots of the equation

$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$

I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn’t work out correctly. This question was in my weekly class worksheet as were this and this question which I previously asked.

Any help will be appreciated.

Solutions Collecting From Web of "Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$"

We have
$$x^2+7x+12-(x+1) \sqrt{x+2} – (x+6)\sqrt{x+7}=0$$
Multiplying the both sides by $3$ gives
$$3x^2+21x+36-3(x+1) \sqrt{x+2} – 3(x+6)\sqrt{x+7}=0\tag1$$
Now since
we have, from $(1)$,
$$(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)-3(x+1) \sqrt{x+2} – 3(x+6)\sqrt{x+7}=0$$
Rearranging terms
$$(x+1)(x+4-3\sqrt{x+2})+(x+6)\sqrt{x+7}\ (\sqrt{x+7}-3)+(x-2)(x+5)=0,$$
$$(x+1)\cdot\frac{(x-2)(x+1)}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}\ (x-2)}{\sqrt{x+7}+3}+(x-2)(x+5)=0$$
and so
We know that $f(x)$ is positive because of $x\ge -2$.

Thus, $\color{red}{x=2}$ is the only solution.

Let $\sqrt{x+2}=a$ and $\sqrt{x+7}= b$.

So, $x + 1 = a^2 -1$

$x+6 = b^2 -1$

$ x^2 + 7x + 12= (x+3)(x+4)= (b^2-4)(a^2+2)$

Substituting and solving the system of equations.