Show that no values of $k$ can make curve intersect the line

Question

The actual question is:

Show that there are no values of $k$ for which the line $y=3-2x$ can intersect the curve $y=2x^2+kx+3$. My approach was:

$$2x^2+kx+3=3-2x\\ \implies2x^2+(k+2)x=0$$

Now no values of k can make $y=3-2x$ intersect $y=2x^2+kx+3$ means

$$(k+2)^2-4\cdot2\cdot0<0$$

or, $(k+2)(k+2)<0$

So there are no values for $k$.

But

If i plot the graph of $y=2x^2-2x+3=0$ and $y=3-2x$, where $k=-2$ then i see them intersect at $(0,3)$ as i also get $(0,3)$ by solving simultaneously.

I cannot get what was the problem in the method of solving?

Answer 1

The discriminant of $$2x^2+(k+2)x$$ is equal to $$(k+2)^2$$ which is always positive. This is why the equation $$2x^2+(k+2)x = 0$$ always has at least one solution (i.e., $x=0$), and if $k\neq -2$, then it has two distinct solutions (the other one being $-\frac{2}{k+2}$)

Answer 2

With $x=0$
$2x^2+kx+3=3$
$3-2x=3$

So $(0,3)$ is an intersection.

So any $k$ yields at least one intersection, and so your method of solving is not wrong.

Answer 3

The equation $2x^2+(k+2)x=0$ has a solution $x=0$ for every $k$. Therefore, it is not amazing the the curves do intersect when $x=0$.