1. **State the problem:** Find the coordinates of points A and B where the curve $y = \sqrt{3x - 2}$ intersects the line $y = \frac{1}{2}x + 1$. 2. **Set the equations equal to find intersection points:** $$\sqrt{3x - 2} = \frac{1}{2}x + 1$$ 3. **Square both sides to eliminate the square root:** $$3x - 2 = \left(\frac{1}{2}x + 1\right)^2$$ 4. **Expand the right side:** $$3x - 2 = \frac{1}{4}x^2 + x + 1$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = \frac{1}{4}x^2 + x + 1 - 3x + 2$$ $$0 = \frac{1}{4}x^2 - 2x + 3$$ 6. **Multiply through by 4 to clear the fraction:** $$0 = x^2 - 8x + 12$$ 7. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-8$, $c=12$. Calculate the discriminant: $$\Delta = (-8)^2 - 4 \times 1 \times 12 = 64 - 48 = 16$$ Calculate roots: $$x = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ So, $$x_1 = \frac{8 + 4}{2} = 6$$ $$x_2 = \frac{8 - 4}{2} = 2$$ 8. **Find corresponding y-values by substituting back into the line equation:** For $x=6$: $$y = \frac{1}{2} \times 6 + 1 = 3 + 1 = 4$$ For $x=2$: $$y = \frac{1}{2} \times 2 + 1 = 1 + 1 = 2$$ 9. **Check the points satisfy the original curve equation:** For $(6,4)$: $$y = \sqrt{3 \times 6 - 2} = \sqrt{18 - 2} = \sqrt{16} = 4$$ For $(2,2)$: $$y = \sqrt{3 \times 2 - 2} = \sqrt{6 - 2} = \sqrt{4} = 2$$ Both points satisfy the curve. **Final answer:** Points of intersection are $A = (2, 2)$ and $B = (6, 4)$.