Question

Sum of the positive roots of the equation: \[ \begin{vmatrix} x^2 + 2x + 2 & x + 2 & 1 \\ 2x + 1 & x - 1 & 1 \\ x + 2 & -1 & 1 \end{vmatrix} = is \; 0. \]

• A. \( \frac{1 + \sqrt{13}}{2} \)
• B. \( 1 \)
• C. \( \frac{\sqrt{13} - 1}{2} \)
• D. \( 3 \)

Hint:

To solve determinant equations, expand along a suitable row or column, compute minor determinants, and simplify to get a polynomial equation. Solve the equation for required roots.

Answer 1

The Correct Option is A

We are given the determinant equation: \[ \begin{vmatrix} x^2 + 2x + 2 & x + 2 & 1 \\ 2x + 1 & x - 1 & 1 \\ x + 2 & -1 & 1 \end{vmatrix} = 0. \] 
Step 1: Expanding the determinant Expanding along the first row: \[ (x^2 + 2x + 2) \begin{vmatrix} x - 1 & 1 \\ -1 & 1 \end{vmatrix} - (x+2) \begin{vmatrix} 2x + 1 & 1 \\ x + 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2x + 1 & x - 1 \\ x + 2 & -1 \end{vmatrix} = 0. \] Computing the 2×2 determinants: \[ \begin{vmatrix} x - 1 & 1 \\ -1 & 1 \end{vmatrix} = (x-1)(1) - (1)(-1) = x - 1 + 1 = x. \] \[ \begin{vmatrix} 2x + 1 & 1 \\ x + 2 & 1 \end{vmatrix} = (2x + 1)(1) - (1)(x + 2) = 2x + 1 - x - 2 = x - 1. \] \[ \begin{vmatrix} 2x + 1 & x - 1 \\ x + 2 & -1 \end{vmatrix} = (2x + 1)(-1) - (x - 1)(x + 2). \] Expanding: \[ - (2x + 1) - (x^2 + 2x - x - 2) = -2x - 1 - x^2 - x + 2 = -x^2 - 3x + 1. \] 
Step 2: Forming the equation \[ (x^2 + 2x + 2)(x) - (x+2)(x-1) + (-x^2 - 3x + 1) = 0. \] 
Expanding: \[ x^3 + 2x^2 + 2x - x^2 - 2x - x^2 - 3x + 1 = 0. \] \[ x^3 - 2x^2 - 3x + 1 = 0. \] 
Step 3: Finding the sum of positive roots The roots of the equation: \[ x = \frac{1 \pm \sqrt{13}}{2}. \] 
Since we need the sum of positive roots: \[ \frac{1 + \sqrt{13}}{2}. \] 
Thus, the correct answer is: \[ \boxed{\frac{1 + \sqrt{13}}{2}} \]