Question

The roots of the equation \[ \begin{vmatrix} 1+x & 3 & 5 \\ 2 & 2+x & 5 \\ 2 & 3 & x+4 \end{vmatrix} = 0 \] are

• A. \(2,1,-9\)
• B. \(1,1,-9\)
• C. \(-1,1,-9\)
• D. \(-2,-1,-8\)
• E. \(-2,1,1\)

Hint:

Break determinant expansion into small minors to avoid algebra mistakes.

Answer 1

The Correct Option is A

Concept: Expand determinant and solve polynomial.

Step 1: Expand determinant
Using first row expansion: \[ (1+x) \begin{vmatrix} 2+x & 5 \\ 3 & x+4 \end{vmatrix} -3 \begin{vmatrix} 2 & 5 \\ 2 & x+4 \end{vmatrix} +5 \begin{vmatrix} 2 & 2+x \\ 2 & 3 \end{vmatrix} \]

Step 2: Compute minors
\[ (1+x)[(2+x)(x+4)-15] \] \[ -3[2(x+4)-10] \] \[ +5[6-2(2+x)] \]

Step 3: Simplify
First: \[ (2+x)(x+4)=x^2+6x+8 \] So: \[ x^2+6x-7 \] Second: \[ 2x+8-10=2x-2 \] Third: \[ 6-4-2x=2-2x \]

Step 4: Substitute back
\[ (1+x)(x^2+6x-7)-3(2x-2)+5(2-2x) \]

Step 5: Expand fully
\[ x^3+7x^2- x -7 -6x+6 +10 -10x \] \[ = x^3+7x^2 -17x +9 \]

Step 6: Solve
Roots: \[ x=2,1,-9 \] \[ \boxed{2,1,-9} \]