Solving $|2x+8|^2 -|9x+36|-9=0$

Question

it looks easy but I messed up something with steps.

$$|2x+8|^2 -|9x+36|-9=0$$

This is what I got

$$|2(x+4)|^2 - | 9(x+4)| - 9 = 0$$

Then I used $u = (x+4)$ and here where it got complicated.

Answer 1

Rather than letting $u = x+4$, you could instead write $$|2(x+4)|^2 - |9(x+4)| - 9 = 2^2|x+4|^2 - 9|x+4| - 9 = 4u^2 - 9u - 9,$$ where $u = |x+4|$. This quadratic factors:

$$4u^2 - 9u - 9 = (u-3)(4u + 3).$$ Alternatively, we can complete the square, or we can simply apply the quadratic formula. In any case, this yields $$u \in \left\{3, -\frac{3}{4}\right\}$$ but only one of these is a solution, since $u = |x+4|$ implies $u \ge 0$. This gives $$|x+4| = 3$$ which in turn implies $x \in \{-1, -7\}$.

Answer 2

Using the hint I gave you: $$4u^2-9u+9=0\,\,\, u\geq 0\,\,\implies u=\frac{9\pm\sqrt{9^2+16\cdot 9}}{8}\implies u=3 \lor u=-\frac{3}{4}$$

Also: $$4u^2+9u+9=0\,\,\ u\leq 0\implies ,\implies u=\frac{-9\pm\sqrt{9^2+16\cdot 9}}{8}\implies u=\frac{3}{4}\lor u=-3$$

So, the solutions are: $x=-1\lor x=-7$.

Answer 3

Here is an alternative method when working with equations involving absolute values.

It consists in replacing $|x|$ by $\, \sigma x\, $ where $\sigma=\pm 1$, however if it simplifies the calculation by avoiding the sign discussions (i.e. $x$ belongs to which or which interval), it also often produces spurious solutions, so we need to plug the obtained values into the initial equation and see if they actually work.

$\begin{align}|2x+8|^2-|9x+36|-9 &=(2x+8)^2-\sigma(9x+36)-9\\ &=4x^2+(32-9\sigma)x+(55-36\sigma)\\ \end{align}$

$\Delta=144+81\overbrace{\sigma^2}^{=1}=225=15^2$

The roots of the quadratic are $x=\dfrac{-32+9\sigma\pm 15}{8}=-7,\ -\frac{19}4,\ -\frac{13}4,\ -1$

And two of them $-1,-7$ are solutions, the other two have to be discarded.

Of course here, there was a simplification since $|x+4|$ could be factored out, but you can learn this method for when you'll have to face equations with no apparent simplification.