Tangent Curve A36Bd0

1. **State the problem:** We need to find the value of $k$ such that the line $y = 3x - 4$ is tangent to the curve $y = (2k - 3)x^2 - kx - (k - 2)$. 2. **Set the line equal to the curve:** Since the line is tangent, the quadratic equation formed by equating the two expressions for $y$ will have exactly one solution (a repeated root). $$3x - 4 = (2k - 3)x^2 - kx - (k - 2)$$ 3. **Rearrange to form a quadratic equation:** $$(2k - 3)x^2 - kx - (k - 2) - 3x + 4 = 0$$ Simplify the linear and constant terms: $$(2k - 3)x^2 - (k + 3)x + (4 - (k - 2)) = 0$$ $$ (2k - 3)x^2 - (k + 3)x + (6 - k) = 0 $$ 4. **Condition for tangency:** The quadratic has exactly one solution if its discriminant $\Delta$ is zero. $$\Delta = b^2 - 4ac = 0$$ Here, $a = 2k - 3$, $b = -(k + 3)$, $c = 6 - k$. Calculate the discriminant: $$\Delta = (-(k + 3))^2 - 4(2k - 3)(6 - k) = (k + 3)^2 - 4(2k - 3)(6 - k)$$ 5. **Expand and simplify:** $$(k + 3)^2 = k^2 + 6k + 9$$ $$(2k - 3)(6 - k) = 12k - 2k^2 - 18 + 3k = -2k^2 + 15k - 18$$ Multiply by 4: $$4(-2k^2 + 15k - 18) = -8k^2 + 60k - 72$$ 6. **Set discriminant to zero:** $$k^2 + 6k + 9 - (-8k^2 + 60k - 72) = 0$$ $$k^2 + 6k + 9 + 8k^2 - 60k + 72 = 0$$ $$9k^2 - 54k + 81 = 0$$ 7. **Simplify the quadratic:** Divide entire equation by 9: $$k^2 - 6k + 9 = 0$$ 8. **Solve for $k$:** $$\Delta_k = (-6)^2 - 4 \times 1 \times 9 = 36 - 36 = 0$$ Since discriminant is zero, one solution: $$k = \frac{6}{2} = 3$$ **Final answer:** $$\boxed{3}$$