Question:
If $x=3$ is a solution of the equation $3 x^{2}+(k-1) x+9=0$, then $k=?$
(a) 11
(b) −11
(c) 13
(d) −13
Solution:
(b) −11
It is given that $x=3$ is a solution of $3 x^{2}+(k-1) x+9=0$; therefore, we have:
$3(3)^{2}+(k-1) \times 3+9=0$
$\Rightarrow 27+3(k-1)+9=0$
$\Rightarrow 3(k-1)=-36$
$\Rightarrow(k-1)=-12$
$\Rightarrow k=-11$