How can I find m using the discriminant? There is a curve $$y=2x^2-4x+8$$ and a line $$y=mx$$.
I have to find the values of m for which the line is tangent to the curve.
I made them equal, then put everything to one side:
$$2x^2-4x+8-mx=0$$
$$2x^2-(4x-mx)+8=0$$
Now, a tangent only touches the curve at one point, so I used the $$b^2-4ac=0$$ discriminant equation, but then here is where I'm stuck. I got:
$$-(4x-mx)^2-4*2*8=0 \tag1 $$
$$-(4x-mx)^2-64=0$$
But then I will get 2 variables again, and $x^2$ and $m^2$ so I don't know how to solve it from here.
 A: $x$ need be there in 1) once again. Else ok as it leads to a quadratic in $m$.
A: So $2x^2−4x+8−mx=0$
Then you need to organize it as
$$2x^2−(4+m)x+8=0$$
Notice that $b = -(4+m)$, but not $-(4x + mx)$ as you listed in your question.
$$b^2 - 4ac=[-(4+m)]^2 -4 \times 2 \times 8 = 0$$
So there is only $m$, no $x$. And you get the point $x$ once you solve $m$ such that the quadratic function about $x$ only have one solution.
A: You have made a couple mistakes:

Mistake 1:
$$2x^2-4x+8-mx=0 \not\Rightarrow 2x^2-(4x-mx)+8=0$$
However, it is correct that:
$$2x^2-4x+8-mx=0 \implies 2x^2-(4x\color{red}{+}mx)+8=0 \tag{1}$$

Mistake 2:
If the form of the quadratic is $ax^2+bx+c=0$ where $a,b$ and $c$ are real constants, then the discriminant $\Delta$ is equal to:
$$\Delta=b^2-4ac$$
Therefore, your $b$ should not contain any $x$ terms!
To avoid this issue, factorize equation $(1)$ like this:
$$2x^2-(4+m)x+8=0$$
Therefore, we can clearly see that $b=-(4+m)$ and thus $b^2=(-1)^2\cdot (4+m)^2=(4+m)^2$.

The correction:
Putting this all together, you should be solving:
$$(4+m)^2-4\cdot 2\cdot 8=0$$
Solving for $m$ gives the two values of $m$ for which the line is tangent to the curve.

To intuitively verify your answer, you can move the slider for $m$ on this Desmos Graph.
