Solve the Equation

Solve the following equation:

\[ \frac{x^3+6x^2+15x+13}{x^2+4x+7} = \sqrt{x+9} - \frac{1}{x+12}. \]

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Solution

Step 1. Determine the Domain

Since the square root is defined only when

\[ x+9\ge0, \]

we must have

\[ x\ge-9. \]

Next, rewrite the numerator:

\[ x^3+6x^2+15x+13 = (x+2)(x^2+4x+7)-1. \]

Therefore,

\[ \frac{x^3+6x^2+15x+13}{x^2+4x+7} = (x+2)-\frac{1}{x^2+4x+7}. \]

Notice that

\[ x^2+4x+7=(x+2)^2+3, \]

and similarly,

\[ x+12=(\sqrt{x+9})^2+3. \]

Hence the equation becomes

\[ (x+2)-\frac{1}{(x+2)^2+3} = \sqrt{x+9} - \frac{1}{(\sqrt{x+9})^2+3}. \tag{1} \]

Step 2. Define a Function

Consider the function

\[ f(u)=u-\frac{1}{u^2+3}, \qquad u\in\mathbb{R}. \]

Its derivative is

\[ f'(u) = 1+\frac{2u}{(u^2+3)^2}. \]

Combining the terms gives

\[ f'(u) = \frac{(u^2+3)^2+2u}{(u^2+3)^2} = \frac{u^4+6u^2+2u+9}{(u^2+3)^2}. \]

Observe that

\[ u^4+6u^2+2u+9 = u^4+5u^2+8+(u+1)^2>0 \]

for every real number \(u\). Therefore,

\[ f'(u)>0, \]

which means that \(f(u)\) is strictly increasing on \(\mathbb{R}\).

Step 3. Solve the Equation

Equation (1) can be written as

\[ f(x+2)=f(\sqrt{x+9}). \]

Since \(f\) is strictly increasing, it is one-to-one. Hence,

\[ x+2=\sqrt{x+9}. \]

Squaring both sides,

\[ (x+2)^2=x+9, \]

which simplifies to

\[ x^2+3x-5=0. \]

Thus,

\[ x=\frac{-3\pm\sqrt{29}}{2}. \]

Using the domain condition \(x\ge-9\) and checking against the original equation:

• \[ \frac{-3-\sqrt{29}}{2} \] does not satisfy \[ x+2=\sqrt{x+9}, \] so it is rejected.
• \[ \frac{-3+\sqrt{29}}{2} \] satisfies the equation.

Final Answer

\[ \boxed{ x=\frac{\sqrt{29}-3}{2} } \]