If \( x=2 \) is a root of \( \begin{vmatrix} 7 & 6 & x\\ x & 3 & 7\\ 2 & x & 2\end{vmatrix}=0 \), then the sum of the other two roots is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −2
Expanding the determinant along the first row:
\( \displaystyle 7(6-7x)-6(2x-14)+x(x^{2}-6) \)
\( \displaystyle =42-49x-12x+84+x^{3}-6x \)
\( \displaystyle =x^{3}-67x+126 \)
There is no \( x^{2} \) term, so the sum of the three roots is 0.
One root is 2, so the other two add up to −2.
Shortcut: Dividing out the known root:
\( \displaystyle x^{3}-67x+126=(x-2)(x-7)(x+9) \)
The other roots are 7 and −9, with sum −2.