Trying to solve $|2x-15| = -x^2 - 5x -8$

Question

My first instinct was to take the positive and negative of the right hand side, resulting in $2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha. Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.

Answer 1

The "is $2x - 15$ positive or negative" dichotomy doesn't work in $\Bbb{C}$. What we can say is that, if $-x^2 - 5x - 8 = |2x - 15|$, then $-x^2 - 5x - 8$ must be a non-negative real number, let's call it $k$. Let's start by solving $$-x^2 - 5x - 8 = k$$ for $x \in \Bbb{C}$. Using the quadratic formula, $$x = \frac{-5 \pm \sqrt{5^2 - 4(8 + k)}}{2} = \frac{-5 \pm i\sqrt{7 + 4k}}{2}.$$ Next, we need $|2x - 15|$ to be equal to this $k$. In particular, $$k = \left|2\frac{-5 \pm i\sqrt{7 + 4k}}{2} - 15\right| = \left|-20 \pm i\sqrt{7 + 4k}\right| = \sqrt{407 + 4k}.$$ Thus, we get a real equation: $$k^2 - 4k - 407 = 0, \quad k \ge 0.$$ This has a unique solution (bearing in mind $k \ge 0$): $$k = 2 + \sqrt{411}.$$ Thus, our solutions for $x \in \Bbb{C}$ must be: $$x = \frac{-5 \pm i\sqrt{7 + 4k}}{2} = \frac{-5 \pm i\sqrt{15 + 4\sqrt{411}}}{2}.$$

Answer 2

I would recommend you to notice that \begin{align*} x^{2} + 5x + 8 = \left(x^{2} + 5x + \frac{25}{4}\right) + \frac{7}{4} = \left(x + \frac{5}{2}\right)^{2} + \frac{7}{4} \geq \frac{7}{4} \end{align*}

Hence we conclude that $-x^{2} - 5x - 8 < 0$ for every possible value of $x$.

Based on such considerations, we conclude the proposed equation has no solutions in $\mathbb{R}$.

Hopefully this helps!

Answer 3

It's easy to see by completing the square that the quadratic is always negative. The two purported real solutions you obtain by setting $2x-15=--x^2-5x-8$ both will fall in the range where $2x-15 \lt 0$, so $2x-15 \neq \vert 2x-15 \vert$ and those solutions are spurious.

Typically, using the letter $x$ (instead of $z$) means that you're working in $\Bbb R$ rather than in $\Bbb C$. If you were working in $\Bbb C$, it wouldn't necessarily be the case that $\vert 2z-15 \vert = \pm(2x-15)$, so it becomes a more complex (pun intended) problem. You would instead break up $z$ into its real and imaginary parts $(z=x+iy)$ and work from there, noting that forcing $-z^2-5z-8$ to be a non-negative real number is a very strong constraint.

Answer 4

For complex numbers, the expression $ \ |2x - 15| \ $ is interpreted as the modulus of the number(s) in brackets. If we use $ \ z \ = \ a + bi \ \ , \ \ a \ , \ b \ $ real, the given equation becomes $$ \sqrt{(2a - 15)^2 \ + \ (2b)^2 } \ \ = \ \ -(a + bi)^2 \ - \ 5(a + bi) \ - \ 8 $$ $$ \Rightarrow \ \ \sqrt{4a^2 - 60a + 225 + 4b^2 } \ \ = \ \ -(a^2 - b^2 + 5a + 8) \ - \ (2ab + 5b)i \ \ . $$

Since the modulus is a real number, the imaginary part of the expression on the right side of this equation must be zero. So we have $ \ 2ab + 5b \ = \ 0 \ \Rightarrow \ a \ = \ -\frac52 \ \ \text{or} \ \ b \ = \ 0 \ \ . $ Since it is easy to show that the "downward-opening" parabola $ \ y \ = \ -x^2 - 5x - 8 \ = \ -\left(x + \frac52 \right)^2 - \frac74 \ \ $ has its vertex "below" the $ \ x-$axis, there can be no purely real solutions to the original equation (as you also found from a graph).

Inserting $ \ a \ = \ -\frac52 \ \ $ into the equation instead leads to $$ \ \sqrt{25 + 150 + 225 + 4b^2 } \ \ = \ \ -(\frac{25}{4} - b^2 - \frac{25}{2} + 8) \ \ \Rightarrow \ \ 2·\sqrt{100 + b^2 } \ \ = \ \ b^2 - \frac{7}{4} $$ $$ \Rightarrow \ \ b^4 \ - \ \frac{15}{2}·b^2 \ - \ \frac{6351}{16} \ \ = \ \ 0 \ \ , $$ a biquadratic equation with two real solutions for $ \ b \ $ . The two complex roots of the original equation are thus $ \ z \ = \ -\frac52 \ \pm \ i·\frac{ \sqrt{15 \ + \ 4 \sqrt{411}}}{2} \ \ . $

Answer 5

The LHS is always non-negative, but the RHS is negative for real $\,x\,$, so no real solutions exist. Let $\,2x-15=z \in \mathbb C \setminus \mathbb R\,$, then substituting $\,x = (z+15)/2\,$ in the original equation:

$$ z^2 + 40 z + 407 = -4 |z| \tag{1} $$

Taking complex conjugates and subtracting $\,(1) - \overline{(1)}\,$:

$$ \require{cancel} \begin{align} (z^2 + 40 z + \cancel{407}) - (\bar z^2 + 40 \bar z + \cancel{407}) &= \bcancel{-4 |z|} + \bcancel{4 |\bar z|} \\ \iff\quad\quad (z- \bar z)(z+\bar z + 40) &= 0 \tag{2} \end{align} $$

The first factor is non-zero $\,z - \bar z \ne 0\,$ because $\,z \not \in \mathbb R\,$, which leaves:

$$ z + \bar z = -40 \tag{3} \quad\iff\quad \text{Re}(z) = -20 $$

Substituting $\,40 = -(z+\bar z)\,$ from $\,(3)\,$ back in $\,(1)\,$:

$$ \cancel{z^2} - (\cancel{z} + \bar z) z + 407 = -4 |z| \;\;\iff\;\; |z|^2 - 4 |z| - 407 = 0 \tag{4} $$

The latter is a quadratic in $\,|z|\,$ with the only positive root $\,|z| = 2 + \sqrt{411}\,$. Then:

$$ \text{Im}(z) = \pm \sqrt{|z|^2 - \text{Re}^2(z)} = \pm \sqrt{\left(2 + \sqrt{411}\right)^2 - (-20)^2} = \pm \sqrt{15 + 4 \sqrt{411}} \tag{5} $$

It follows from $\,(3)\,$ and $\,(5)\,$ that:

$$ z = \text{Re}(z) + i\,\text{Im}(z) = -20 \pm i\sqrt{15 + 4 \sqrt{411}} \\ \implies\quad\;\; x = \frac{z+15}{2} = \frac{-5 \pm i\sqrt{15 + 4 \sqrt{411}}}{2} \;\; $$