1. **State the problem:** We have a curve defined by the equation $$y = (2k - 3)x^2 - kx - (k - 2)$$ where $k$ is a constant. We are told the line $$y = 3x - 4$$ is tangent to this curve. We need to find the value of $k$. 2. **Recall the tangent condition:** A line is tangent to a curve if they touch at exactly one point. This means the quadratic equation formed by setting the curve equal to the line has exactly one solution (a repeated root). The discriminant of this quadratic must be zero. 3. **Set the curve equal to the line:** $$ (2k - 3)x^2 - kx - (k - 2) = 3x - 4 $$ 4. **Bring all terms to one side:** $$ (2k - 3)x^2 - kx - (k - 2) - 3x + 4 = 0 $$ Simplify the linear and constant terms: $$ (2k - 3)x^2 - (k + 3)x + (-k + 2) = 0 $$ 5. **Identify coefficients for the quadratic in $x$:** $$ a = 2k - 3 $$ $$ b = -(k + 3) $$ $$ c = -k + 2 $$ 6. **Apply the discriminant condition for tangency:** $$ \Delta = b^2 - 4ac = 0 $$ Substitute $a$, $b$, and $c$: $$ (-(k + 3))^2 - 4(2k - 3)(-k + 2) = 0 $$ Simplify: $$ (k + 3)^2 - 4(2k - 3)(-k + 2) = 0 $$ 7. **Expand terms:** $$ (k + 3)^2 = k^2 + 6k + 9 $$ $$ (2k - 3)(-k + 2) = 2k(-k + 2) - 3(-k + 2) = -2k^2 + 4k + 3k - 6 = -2k^2 + 7k - 6 $$ 8. **Substitute back:** $$ k^2 + 6k + 9 - 4(-2k^2 + 7k - 6) = 0 $$ 9. **Distribute the -4:** $$ k^2 + 6k + 9 + 8k^2 - 28k + 24 = 0 $$ 10. **Combine like terms:** $$ 9k^2 - 22k + 33 = 0 $$ 11. **Solve the quadratic equation for $k$:** Use the quadratic formula: $$ k = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 9 \times 33}}{2 \times 9} $$ Calculate the discriminant inside the square root: $$ 484 - 1188 = -704 $$ Since the discriminant is negative, no real solutions exist for $k$ here. This suggests a mistake in simplification. 12. **Re-examine step 9:** The sign in step 9 should be carefully checked: $$ k^2 + 6k + 9 - 4(-2k^2 + 7k - 6) = k^2 + 6k + 9 + 8k^2 - 28k + 24 $$ This is correct. 13. **Combine terms again:** $$ 9k^2 - 22k + 33 = 0 $$ 14. **Check discriminant again:** $$ \Delta = (-22)^2 - 4 \times 9 \times 33 = 484 - 1188 = -704 $$ Negative discriminant means no real $k$. 15. **Check if $a = 0$ (special case):** If $a = 2k - 3 = 0$, then $k = \frac{3}{2} = 1.5$. Substitute $k=1.5$ into the quadratic: $$ a = 0, b = -(1.5 + 3) = -4.5, c = -1.5 + 2 = 0.5 $$ Equation becomes: $$ 0 \times x^2 - 4.5x + 0.5 = 0 $$ This is linear, so it has one solution, meaning the line is tangent. 16. **Conclusion:** The value of $k$ for which the line is tangent to the curve is: $$ \boxed{\frac{3}{2}} $$