1. Problem 49: Find the number of intersection points between $y = \sqrt{|x|}$ and $5y = x + 6$. 2. Problem 50: Find the number of intersection points between $y = |x^2 - 4|$ and $y = \frac{x^2}{2} + 4$. --- ### Problem 49 1. State the problem: Find intersections of $y = \sqrt{|x|}$ and $5y = x + 6$. 2. Express $y$ from the second equation: $y = \frac{x + 6}{5}$. 3. Set equal: $\sqrt{|x|} = \frac{x + 6}{5}$. 4. Square both sides: $|x| = \left(\frac{x + 6}{5}\right)^2 = \frac{(x + 6)^2}{25}$. 5. Multiply both sides by 25: $25|x| = (x + 6)^2$. 6. Consider two cases for $|x|$: - Case 1: $x \geq 0$, then $25x = (x + 6)^2$. - Case 2: $x < 0$, then $-25x = (x + 6)^2$. 7. Solve Case 1: $25x = x^2 + 12x + 36$. Rearrange: $x^2 - 13x + 36 = 0$. Factor or use quadratic formula: $x = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2}$. Solutions: $x = 9$ or $x = 4$ (both $\geq 0$ valid). 8. Solve Case 2: $-25x = x^2 + 12x + 36$. Rearrange: $x^2 + 37x + 36 = 0$. Use quadratic formula: $x = \frac{-37 \pm \sqrt{37^2 - 4 \cdot 36}}{2} = \frac{-37 \pm \sqrt{1369 - 144}}{2} = \frac{-37 \pm \sqrt{1225}}{2} = \frac{-37 \pm 35}{2}$. Solutions: $x = -1$ or $x = -36$ (both $< 0$ valid). 9. Total valid $x$ values: $-36, -1, 4, 9$. 10. Find corresponding $y$ values using $y = \frac{x + 6}{5}$: - For $x = -36$, $y = \frac{-36 + 6}{5} = \frac{-30}{5} = -6$ (Check $y = \sqrt{|x|} = \sqrt{36} = 6$, mismatch in sign, discard). - For $x = -1$, $y = \frac{-1 + 6}{5} = 1$ (Check $\sqrt{1} = 1$, valid). - For $x = 4$, $y = \frac{4 + 6}{5} = 2$ (Check $\sqrt{4} = 2$, valid). - For $x = 9$, $y = \frac{9 + 6}{5} = 3$ (Check $\sqrt{9} = 3$, valid). 11. Valid intersection points: $( -1, 1 ), (4, 2), (9, 3)$. 12. Number of intersection points: **3**. --- ### Problem 50 1. State the problem: Find intersections of $y = |x^2 - 4|$ and $y = \frac{x^2}{2} + 4$. 2. Set equal: $|x^2 - 4| = \frac{x^2}{2} + 4$. 3. Consider two cases for absolute value: - Case 1: $x^2 - 4 \geq 0 \Rightarrow |x^2 - 4| = x^2 - 4$. - Case 2: $x^2 - 4 < 0 \Rightarrow |x^2 - 4| = -(x^2 - 4) = 4 - x^2$. 4. Case 1 ($x^2 \geq 4$): $x^2 - 4 = \frac{x^2}{2} + 4$. Rearrange: $x^2 - \frac{x^2}{2} = 4 + 4$. $\frac{x^2}{2} = 8$. $x^2 = 16$. So $x = \pm 4$. Check condition $x^2 \geq 4$ holds. 5. Case 2 ($x^2 < 4$): $4 - x^2 = \frac{x^2}{2} + 4$. Rearrange: $4 - x^2 - \frac{x^2}{2} = 4$. $-\frac{3x^2}{2} = 0$. $x^2 = 0$. So $x = 0$. Check condition $x^2 < 4$ holds. 6. Intersection points at $x = -4, 0, 4$. 7. Find $y$ values: - At $x = -4$: $y = \frac{16}{2} + 4 = 8 + 4 = 12$. - At $x = 0$: $y = \frac{0}{2} + 4 = 4$. - At $x = 4$: $y = 12$. 8. Number of intersection points: **3**. --- Final answers: - Problem 49: 3 intersection points. - Problem 50: 3 intersection points.