Okay, so I'm self studying my A level mathematics and I don't have a tutor to ask from so I hope someone can help me here.
My question is:
Q: The equation of curve is $y=\dfrac{12}x$ and the of a line $L$ is $2 x + y= k$ where $k$ is a constant. Find the set of values of $k$ for which $L$ does not intersect the curve.
So here's my answer:
Ans:
•Solving simultaneously,
$$\frac{12}x = kx - 2x^2$$
$$2x^2 - kx + 12 = 0$$
•Since $L$ doesn't intersect there are no real roots, therefore the discriminant is
$$b^2 - 4ac < 0$$
•substituting values
$$(-k)^2 - 4(2)(12) < 0$$
$$k^2 - 96 < 0$$
$$(k - \sqrt{96})(k + \sqrt{96}) < 0$$
Now I don't know how to get the answer. I know to draw a graph, but for $x$ values, I don't understand how to use it for $k$ or neither do I know any other method.
(Can you keep the explanation simple in A level boundaries?)
Your working so far is fine. Now you have $$(k+\sqrt{96})(k-\sqrt{96})<0$$ To satisfy the inequality, either of the factors must be negative and the other positive. The range that satisfies that is $$-\sqrt{96}<k<\sqrt{96}$$
You can actually just treat $k$ as if it is an $x$. It's just a dummy variable. Don't be too concerned about the 'appearance'.