1. **Problem:** Solve the equation $|x^2 - 14| = 5x$ for $x > 0$. 2. **Formula and approach:** Since the equation involves an absolute value, we consider two cases: - Case 1: $x^2 - 14 = 5x$ - Case 2: $x^2 - 14 = -5x$ 3. **Case 1:** $$x^2 - 14 = 5x$$ Rearrange to standard quadratic form: $$x^2 - 5x - 14 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-5$, $c=-14$: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 1 \times (-14)}}{2} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2}$$ $$x = \frac{5 \pm 9}{2}$$ Possible solutions: - $x = \frac{5 + 9}{2} = 7$ - $x = \frac{5 - 9}{2} = -2$ (discard since $x > 0$) 4. **Case 2:** $$x^2 - 14 = -5x$$ Rearranged: $$x^2 + 5x - 14 = 0$$ Apply quadratic formula with $a=1$, $b=5$, $c=-14$: $$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times (-14)}}{2} = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2}$$ $$x = \frac{-5 \pm 9}{2}$$ Possible solutions: - $x = \frac{-5 + 9}{2} = 2$ - $x = \frac{-5 - 9}{2} = -7$ (discard since $x > 0$) 5. **Final solutions for $x > 0$ are:** $$x = 2 \text{ and } x = 7$$