Question

Find all values of $x$ such that $\dfrac{x}{x+4} = -\dfrac{9}{x+3}$. If you find more than one value, then list your solutions in increasing order, separated by commas.

Answer 1

We start with

$\dfrac{x}{x+4} = -\dfrac{9}{x+3}$

Assuming that $x \notin \{-4,\ -3\}$ we can cross multiply the equation:

$x(x+3) = -9(x+4) \iff x^2+3x=-9x-36 \iff x^2+12x+36=0$

You can recognize the perfect square pattern in the quadratic equation:

$x^2+12x+36 = (x)^2 + 2\cdot x \cdot 6 + (6)^2 = (x+6)^2$

So, we have

$x^2+12x+36=0 \iff (x+6)^2 = 0 \iff x+6=0 \iff x=-6$

Which is the only solution, with multiplicity 2.

Answer 2

Multiplying both sides by $x+4$ and by $x+3$ gives

\[x(x+3) = -9(x+4).\]

Expanding both sides gives $x^2 + 3x = -9x -36$, and rearranging gives $x^2 + 12x + 36=0$. To factor this quadratic, we seek values $A$ and $B$ such that

\[x^2 + 12x + 36 = (x + A)(x + B) = x^2 + (A + B)x + AB,\]

so $A + B = 12$ and $AB = 36$. We find that $A = B = 6$ works, so

\[x^2 + 12x + 36 = (x + 6)(x + 6) = (x + 6)^2.\]

Therefore, the only solution is $x = \boxed{-6}$. We say that this is a "double root" of the quadratic.