1. Stating the problem: Solve the equation $$|x|4x + 3 = -5$$. 2. First, isolate the absolute value term and constants: $$|x|4x = -5 - 3$$ which simplifies to $$|x|4x = -8$$. 3. Note that $$|x|$$ is always non-negative (i.e., $$|x| \geq 0$$), and $$4x$$ depends on the sign of $$x$$. 4. Consider the expression $$|x|4x$$. Since $$|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$$, then: - If $$x \geq 0$$, $$|x|4x = x \cdot 4x = 4x^2$$. - If $$x < 0$$, $$|x|4x = (-x) \cdot 4x = -4x^2$$. 5. Solve for each case: Case 1: $$x \geq 0$$ $$4x^2 = -8$$ Divide both sides by 4: $$x^2 = -2$$ Since $$x^2$$ cannot be negative for real numbers, no real solution here. Case 2: $$x < 0$$ $$-4x^2 = -8$$ Multiply both sides by -1: $$4x^2 = 8$$ Divide both sides by 4: $$x^2 = 2$$ Take square root: $$x = \pm \sqrt{2}$$ But since $$x < 0$$, only $$x = -\sqrt{2}$$ is valid. 6. Final answer: $$x = -\sqrt{2}$$. This is the only real solution to the equation.