If \(x=-4\) is a root of \(\begin{vmatrix}x&2&3\\1&x&1\\3&2&x\end{vmatrix}=0\), then the sum of the other 2 roots is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Expanding the determinant along the first row:
\(x(x^2 - 2) - 2(x - 3) + 3(2 - 3x) = 0\)
\(x^3 - 2x - 2x + 6 + 6 - 9x = 0\)
\(x^3 - 13x + 12 = 0\)
The cubic has no \(x^2\) term, so the sum of its three roots is 0. One root is −4, so the other two add up to
\(0 - (-4) = 4\)
Check by factorising:
\(x^3 - 13x + 12 = (x + 4)(x^2 - 4x + 3)\)
\(x^3 - 13x + 12 = (x + 4)(x - 1)(x - 3)\)
The other roots are 1 and 3, and 1 + 3 = 4.
NCERT Exemplar:
Determinants
