The deriations from the mean of a set of numbers are \( (k-12)^{2}(k+7),-2, k \) a-d \( (k+2)^{2} \), where \( k \) is a constant. Find the value of \( K \) if the mean deviation is \( O \)

To find the value of \( k \) such that the mean deviation is \( 0 \), we first need to understand the concept of mean deviation. The mean deviation is defined as the average of the absolute deviations from the mean. If the mean deviation is \( 0 \), it implies that all deviations must be \( 0 \). Given the deviations from the mean are: 1. \( (k-12)^{2}(k+7) \) 2. \( -2 \) 3. \( k \) 4. \( (k+2)^{2} \) We can set up the equation for the mean deviation being \( 0 \): \[ (k-12)^{2}(k+7) + (-2) + k + (k+2)^{2} = 0 \] Now, let's simplify this equation step by step. 1. Expand \( (k-12)^{2} \): \[ (k-12)^{2} = k^{2} - 24k + 144 \] 2. Expand \( (k+2)^{2} \): \[ (k+2)^{2} = k^{2} + 4k + 4 \] 3. Substitute these expansions back into the equation: \[ (k^{2} - 24k + 144)(k + 7) - 2 + k + (k^{2} + 4k + 4) = 0 \] Now, we will calculate \( (k^{2} - 24k + 144)(k + 7) \) and combine all terms. Let's perform the multiplication and then combine all terms. 1. Calculate \( (k^{2} - 24k + 144)(k + 7) \): \[ = k^{3} + 7k^{2} - 24k^{2} - 168k + 144k + 1008 \] \[ = k^{3} - 17k^{2} - 24k + 1008 \] 2. Now, combine all terms: \[ k^{3} - 17k^{2} - 24k + 1008 - 2 + k + (k^{2} + 4k + 4) = 0 \] \[ = k^{3} - 16k^{2} - 19k + 1010 = 0 \] Now we need to solve the cubic equation \( k^{3} - 16k^{2} - 19k + 1010 = 0 \). Let's use a numerical method or a root-finding algorithm to find the value of \( k \). I'll proceed to find the roots of this cubic equation. Solve the equation(The complex numbers system) by following steps: - step0: Solve for \(k\) in the complex numbers system: \(k^{3}-16k^{2}-19k+1010=0\) - step1: Calculate: \(\begin{align}&k\approx -7.045239\\&k\approx 11.52262+3.253991i\\&k\approx 11.52262-3.253991i\end{align}\) - step2: Rewrite: \(k_{1}\approx 11.52262-3.253991i,k_{2}\approx 11.52262+3.253991i,k_{3}\approx -7.045239\) - step3: Remove the complex number(s): \(k\approx -7.045239\) The value of \( k \) that makes the mean deviation equal to \( 0 \) is approximately \[ k \approx -7.045239. \] This means that when \( k \) is around \(-7.045239\), the deviations from the mean will sum to zero, resulting in a mean deviation of \( 0 \).