Question

If \( \begin{vmatrix} 9 & 25 & 16 16 & 36 & 25 25 & 49 & 36 \end{vmatrix} = K \), then \( K, K + 1 \) are the roots of the equation:

• A. \( x^2 - 13x + 42 = 0 \)
• B. \( x^2 - 15x + 56 = 0 \)
• C. \( x^2 - 19x + 90 = 0 \)
• D. \( x^2 - 17x + 72 = 0 \)

Hint:

Whenever a row or column in a determinant consists entirely of identical non-zero numbers (like \(\begin{bmatrix} 2 & 2 & 2 \end{bmatrix}\)), factor it out immediately to leave a row of ones. Then, apply column differences relative to the last column to generate two zeros in a single step, saving valuable calculation time!

Answer 1

The Correct Option is B

Concept: Determinants containing consecutive perfect squares can be greatly simplified by recognizing their underlying structure. Notice that the elements can be written as: \[ \begin{vmatrix} 3^2 & 5^2 & 4^2 4^2 & 6^2 & 5^2 5^2 & 7^2 & 6^2 \end{vmatrix} \] By applying successive row differences, we can reduce the quadratic expressions (squares) to linear differences, making numerical evaluation straightforward. Once the value of \( K \) is determined, a quadratic equation having roots \(\alpha\) and \(\beta\) is given by: \[ x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \]

Step 1: Applying row operations to reduce the order of the terms inside the determinant.
Let us perform the operations \(R_3 \rightarrow R_3 - R_2\) and \(R_2 \rightarrow R_2 - R_1\): \[ K = \begin{vmatrix} 9 & 25 & 16 16 - 9 & 36 - 25 & 25 - 16 25 - 16 & 49 - 36 & 36 - 25 \end{vmatrix} \] Evaluating the differences: \[ K = \begin{vmatrix} 9 & 25 & 16 7 & 11 & 9 9 & 13 & 11 \end{vmatrix} \]

Step 2: Applying another row operation to create smaller, manageable constants.
Perform the row operation \(R_3 \rightarrow R_3 - R_2\): \[ K = \begin{vmatrix} 9 & 25 & 16 7 & 11 & 9 9 - 7 & 13 - 11 & 11 - 9 \end{vmatrix} \] \[ K = \begin{vmatrix} 9 & 25 & 16 7 & 11 & 9 2 & 2 & 2 \end{vmatrix} \] Take out the common factor \(2\) from the third row (\(R_3\)): \[ K = 2 \begin{vmatrix} 9 & 25 & 16 7 & 11 & 9 1 & 1 & 1 \end{vmatrix} \]

Step 3: Applying column operations to create zeros and evaluating the determinant.
Perform the column operations \(C_1 \rightarrow C_1 - C_3\) and \(C_2 \rightarrow C_2 - C_3\): \[ K = 2 \begin{vmatrix} 9 - 16 & 25 - 16 & 16 7 - 9 & 11 - 9 & 9 1 - 1 & 1 - 1 & 1 \end{vmatrix} \] \[ K = 2 \begin{vmatrix} -7 & 9 & 16 -2 & 2 & 9 0 & 0 & 1 \end{vmatrix} \] Expanding along the third row (\(R_3\)): \[ K = 2 \cdot 1 \cdot \begin{vmatrix} -7 & 9 -2 & 2 \end{vmatrix} \] \[ K = 2 \cdot [(-7)(2) - (9)(-2)] = 2 \cdot [-14 + 18] = 2 \cdot 4 = 7 \]

Step 4: Forming the quadratic equation using the determined roots.
Since \( K = 7 \), the two roots are:
• \(\alpha = K = 7\)
• \(\beta = K + 1 = 7 + 1 = 8\) Now, find the structural components for the quadratic equation:
• \(\text{Sum of roots} = 7 + 8 = 15\)
• \(\text{Product of roots} = 7 \times 8 = 56\) Substituting these into the standard quadratic form: \[ x^2 - 15x + 56 = 0 \] This perfectly matches option (B).