Let the equation $x(x+2)(12-k)=2$ have equal roots. Then the distance of the point $\left(k, \frac{k}{2}\right)$ from the line $3 x+4 y+5=0$ is

<p>Given the equation $ x(x+2)(12-k) = 2 $, we want it to have equal roots. To achieve this, we need to manipulate it into a quadratic form in terms of $ x $:</p> <p>$ x^2 + 2x - \frac{2}{12-k} = 0 $</p> <p>For this quadratic equation to have equal (repeated) roots, the discriminant $ D $ must be zero. The discriminant $ D $ for the equation $ ax^2 + bx + c = 0 $ is given by:</p> <p>$ D = b^2 - 4ac $</p> <p>Substituting $ a = 1 $, $ b = 2 $, and $ c = -\frac{2}{12-k} $ into the discriminant formula, we get:</p> <p>$ 4 - 4\left(-\frac{2}{12-k}\right) = 0 $</p> <p>Simplifying the expression:</p> <p>$ 1 + \frac{2}{12-k} = 0 $</p> <p>Solving for $ k $:</p> <p>$ \frac{2}{12-k} = -1 \quad \Rightarrow \quad 2 = -(12 - k) \quad \Rightarrow \quad 2 = -12 + k \quad \Rightarrow \quad k = 14 $</p> <p>Now, consider the point $(k, \frac{k}{2})$, which becomes $(14, 7)$. We need to find its distance from the line $3x + 4y + 5 = 0$. The formula for the distance $ d $ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:</p> <p>$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $</p> <p>Substituting $ (x_1, y_1) = (14, 7) $ and the line coefficients $ A = 3 $, $ B = 4 $, $ C = 5 $:</p> <p>$ d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}} $</p> <p>Calculating the numerator:</p> <p>$ 3 \times 14 + 4 \times 7 + 5 = 42 + 28 + 5 = 75 $</p> <p>And the denominator:</p> <p>$ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $</p> <p>Therefore, the distance is:</p> <p>$ d = \frac{75}{5} = 15 $</p> <p>Thus, the distance is 15.</p>