A curve is defined parametrically by the equations
$$ x = t^3 - 6t + 4, y = t - 3 + \frac{2}{t} $$
The first question, which I've partially solved, was to find the equations of the normals to the curve at the points where the curve meets the x-axis. The next step is to find the coordinates of their point of intersection.
I've already found one normal:
$$ 0 = t^3 - 6t + 4 \\ 0 = (t-2)(t^2+2t-2) \ $$
So either $t=2$ or $t^2+2t-2 = 0$. How do you take the latter equation and find the second value of $t$? Next we need to find the gradient:
$$\begin{align} \frac{dy}{dx} & = \frac{dy}{dt} \times \frac{dt}{dx} \\ \frac{dy}{dt} & = 1 - \frac{2}{t^2} \\ \frac{dt}{dx} & = \frac{1}{3t^2-6} \end{align}$$
Taking $t = 2$, then:
$$ \frac{dy}{dx} = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} $$
And so the line of the normal will be $y + 12x = 0$. But I can't find the other normal equation (listed as $y+3x+3=0$). How do I find it?
Note To meet x-axis put $y=0$ not $x$ and hence $t=1,2$
The gradient is $$y'=\frac{t^2-2}{t^2}*\frac{1}{3(t^2-2)}=\frac{1}{3t^2}$$ At $t=1,2$ $$y'=\frac13\implies slope=-3\\y'=\frac1{12}\implies slope=-12$$