Question

Let the mean and the variance of seven observations \(2,4,\alpha,8,\beta,12,14\), \( \alpha < \beta \), be \(8\) and \(16\) respectively. Then the quadratic equation whose roots are \(3\alpha+2\) and \(2\beta+1\) is :

• A. \(x^2-35x+306=0\)
• B. \(x^2-41x+420=0\)
• C. \(x^2-45x+506=0\)
• D. \(x^2-37x+342=0\)

Answer 1

The Correct Option is D

Step 1: Understanding the Question
We are given a set of seven data points, two of which are unknown (\(\alpha, \beta\)). We are also given the mean and variance of this dataset. We need to first find the values of \(\alpha\) and \(\beta\), then form a new quadratic equation whose roots are derived from \(\alpha\) and \(\beta\).
Step 2: Key Formula or Approach
For a set of \(n\) observations \(x_1, x_2, \dots, x_n\):
Mean \( \mu = \frac{\sum_{i=1}^n x_i}{n} \).
Variance \( \sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - \mu^2 \).
A quadratic equation with roots \(r_1\) and \(r_2\) is given by \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \).
Step 3: Detailed Explanation
The seven observations are \(2, 4, \alpha, 8, \beta, 12, 14\). Here \(n=7\).
Given mean \( \mu = 8 \) and variance \( \sigma^2 = 16 \).
Using the mean to find \( \alpha + \beta \):
\[ \mu = \frac{2+4+\alpha+8+\beta+12+14}{7} = 8 \] \[ \frac{40 + \alpha + \beta}{7} = 8 \] \[ 40 + \alpha + \beta = 56 \] \[ \alpha + \beta = 16 \quad \text{(Equation 1)} \] Using the variance to find \( \alpha^2 + \beta^2 \):
\[ \sigma^2 = \frac{\sum x_i^2}{7} - \mu^2 = 16 \] \[ \frac{2^2+4^2+\alpha^2+8^2+\beta^2+12^2+14^2}{7} - 8^2 = 16 \] \[ \frac{4+16+\alpha^2+64+\beta^2+144+196}{7} - 64 = 16 \] \[ \frac{424 + \alpha^2 + \beta^2}{7} = 80 \] \[ 424 + \alpha^2 + \beta^2 = 560 \] \[ \alpha^2 + \beta^2 = 136 \quad \text{(Equation 2)} \] Solving for \( \alpha \) and \( \beta \):
We know that \( (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \).
Using Equations 1 and 2: \[ (16)^2 = 136 + 2\alpha\beta \] \[ 256 = 136 + 2\alpha\beta \] \[ 120 = 2\alpha\beta \implies \alpha\beta = 60 \] Now we have \( \alpha+\beta=16 \) and \( \alpha\beta=60 \). \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( t^2 - 16t + 60 = 0 \).
Factoring the equation: \( (t-10)(t-6) = 0 \). The roots are \(t=6\) and \(t=10\).
Given that \( \alpha<\beta \), we have \( \alpha = 6 \) and \( \beta = 10 \).
Forming the new quadratic equation:
The roots of the new equation are \( r_1 = 3\alpha + 2 \) and \( r_2 = 2\beta + 1 \).
\[ r_1 = 3(6) + 2 = 18 + 2 = 20 \] \[ r_2 = 2(10) + 1 = 20 + 1 = 21 \] Sum of new roots = \( 20 + 21 = 41 \).
Product of new roots = \( 20 \times 21 = 420 \).
The required quadratic equation is:
\[ x^2 - (41)x + 420 = 0 \] Step 4: Final Answer
The resulting equation is \( x^2 - 41x + 420 = 0 \), which corresponds to option (B).