1. **State the problem:** Solve the equation $$x |x - 8| = 24x + 4$$ for $x$ and identify which solutions are roots or extraneous. 2. **Understand the absolute value:** The expression $|x - 8|$ splits into two cases: - Case 1: $x - 8 \geq 0 \Rightarrow |x - 8| = x - 8$ - Case 2: $x - 8 < 0 \Rightarrow |x - 8| = -(x - 8) = 8 - x$ 3. **Solve Case 1 ($x \geq 8$):** $$x |x - 8| = x(x - 8) = x^2 - 8x$$ Set equal to right side: $$x^2 - 8x = 24x + 4$$ Bring all terms to one side: $$x^2 - 8x - 24x - 4 = 0$$ $$x^2 - 32x - 4 = 0$$ Use quadratic formula: $$x = \frac{32 \pm \sqrt{32^2 - 4 \cdot 1 \cdot (-4)}}{2} = \frac{32 \pm \sqrt{1024 + 16}}{2} = \frac{32 \pm \sqrt{1040}}{2}$$ Simplify $$\sqrt{1040} = \sqrt{16 \cdot 65} = 4\sqrt{65}$$: $$x = \frac{32 \pm 4\sqrt{65}}{2} = 16 \pm 2\sqrt{65}$$ Check domain $x \geq 8$: - $16 - 2\sqrt{65} \approx 16 - 16.12 = -0.12$ (not in domain) - $16 + 2\sqrt{65} \approx 16 + 16.12 = 32.12$ (valid) So from Case 1, valid root: $$x = 16 + 2\sqrt{65}$$ 4. **Solve Case 2 ($x < 8$):** $$x |x - 8| = x(8 - x) = 8x - x^2$$ Set equal to right side: $$8x - x^2 = 24x + 4$$ Bring all terms to one side: $$-x^2 + 8x - 24x - 4 = 0$$ $$-x^2 - 16x - 4 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 - 16x - 4 = 0 \Rightarrow \cancel{-}(-x^2 - 16x - 4) = 0 \Rightarrow x^2 + 16x + 4 = 0$$ Use quadratic formula: $$x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 4}}{2} = \frac{-16 \pm \sqrt{256 - 16}}{2} = \frac{-16 \pm \sqrt{240}}{2}$$ Simplify $$\sqrt{240} = \sqrt{16 \cdot 15} = 4\sqrt{15}$$: $$x = \frac{-16 \pm 4\sqrt{15}}{2} = -8 \pm 2\sqrt{15}$$ Check domain $x < 8$: - $-8 - 2\sqrt{15} \approx -8 - 7.75 = -15.75$ (valid) - $-8 + 2\sqrt{15} \approx -8 + 7.75 = -0.25$ (valid) So from Case 2, valid roots: $$x = -8 - 2\sqrt{15}, \quad x = -8 + 2\sqrt{15}$$ 5. **Check all candidate solutions in original equation:** - $x = -8 - 2\sqrt{15}$: Substitute and verify equality (true root). - $x = -8 + 2\sqrt{15}$: Substitute and verify equality (true root). - $x = 16 + 2\sqrt{65}$: Substitute and verify equality (true root). - $x = 16 - 2\sqrt{65}$: Not in domain for Case 1, check substitution: Substitute $x = 16 - 2\sqrt{65}$ into original: Calculate $|x - 8| = |16 - 2\sqrt{65} - 8| = |8 - 2\sqrt{65}|$. Since $2\sqrt{65} \approx 16.12$, $8 - 16.12 = -8.12$, absolute value is $8.12$. Left side: $$x |x - 8| = (16 - 2\sqrt{65}) \times 8.12$$ Right side: $$24x + 4 = 24(16 - 2\sqrt{65}) + 4$$ Numerical approximation shows they are not equal, so extraneous. 6. **List roots in increasing order:** $$-8 - 2\sqrt{15} < -8 + 2\sqrt{15} < 16 - 2\sqrt{65} < 16 + 2\sqrt{65}$$ 7. **Label each:** - $-8 - 2\sqrt{15}$: ROOT - $-8 + 2\sqrt{15}$: ROOT - $16 - 2\sqrt{65}$: EXTRANEOUS - $16 + 2\sqrt{65}$: ROOT **Final answer:** $$\boxed{-8 - 2\sqrt{15} < -8 + 2\sqrt{15} < 16 - 2\sqrt{65} < 16 + 2\sqrt{65}}$$ with labels: ROOT, ROOT, EXTRANEOUS, ROOT