Find $k$ for which $x = 3$ is a solution of the quadratic equation $\left(k + 2\right)x^2 - kx + 6 = 0$. Thus find the other root of the equation.
$\because \;$ $x = 3$ is a solution of the quadratic equation $\left(k + 2\right)x^2 - kx + 6 = 0$, we have
$\left(k + 2\right) \times 3^2 - k \times 3 + 6 = 0$
i.e. $\;$ $9k + 18 - 3k + 6 = 0$
i.e. $\;$ $6k = - 24$ $\implies$ $k = -4$
Substituting the value of $k$, the given quadratic equation becomes,
$\left(-4 + 2\right)x^2 - \left(-4\right)x + 6 = 0$
i.e. $\;$ $2 x^2 - 4 x - 6 = 0$
i.e. $\;$ $x^2 - 2x - 3 = 0$
i.e. $\;$ $x^2 - 3x + x - 3 = 0$
i.e. $\;$ $x \left(x - 3\right) + 1 \left(x - 3\right) = 0$
i.e. $\;$ $\left(x + 1\right) \left(x - 3\right) = 0$
i.e. $\;$ $x + 1 = 0 \;$ or $\;$ $x - 3 = 0$
i.e. $\;$ $x = -1 \;$ or $\;$ $x = 3$
$\therefore \;$ The other root of the equation is $x = -1$.
ALTERNATIVELY
Comparing $\;$ $x^2 - 2x - 3 = 0$ $\;$ with standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives
$a = 1, \; b = -2, \; c = -3$
Given: $\;$ One root of the quadratic equation $= \alpha = 3$
Let the other root of the quadratic equation $= \beta$
Sum of roots $= \alpha + \beta = \dfrac{-b}{a}$
i.e. $\;$ $3 + \beta = \dfrac{2}{1} = 2$
$\implies$ $\beta = -1$
$\therefore \;$ The other root of the equation is $\beta = -1$.