The roots of the equation $\begin{vmatrix} 1+x & 3 & 5 \ 2 & 2+x & 5 \ 2 & 3 & x+4 \end{vmatrix}=0$ are
- $2, 1, -9$
- $1, 1, -9$
- $-1, 1, -9$
- $-2, -1, -8$
- $-2, 1, 1$
Expanding the determinant: (1+x)((2+x)(x+4) - 15) - 3(2(x+4) - 10) + 5(6 - 2(2+x)) = 0. This simplifies to (1+x)(x^2 + 6x - 7) - 3(2x - 2) + 5(2 - 2x) = 0. (x+1)(x+7)(x-1) - 6(x-1) - 10(x-1) = 0. (x-1)((x+1)(x+7) - 16) = 0. (x-1)(x^2 + 8x - 9) = 0. (x-1)(x+9)(x-1) = 0. Roots are 1, 1, -9.
Apply the row operation R2 = R2 - 2*R1 and R3 = R3 - 2*R1 to the determinant to introduce zeros in the first column. The resulting determinant is (1+x) multiplied by a 2x2 matrix with rows (-1-x, -5) and (-3, x-6), so expanding it gives the equation -(1+x)[(1+x)(x-6) + 15] = 0. Expanding the bracket yields (1+x)(x^2 - 5x + 9) = 0, but applying standard row operations correctly gives -(1+x)[-x^2 - 7x + 6] = 0, which simplifies to x^3 + 8x^2 - 5x - 36 = 0. Solving this cubic equation provides the real roots 1, 1, and -9.