How can I find m using the discriminant?

Question

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.

Answer 1

You have made a couple mistakes:

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

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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 $b^2=(-1)^2\cdot (4+m)^2=(4+m)^2$.

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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.

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

Answer 2

$x$ need be there in 1) once again. Else ok as it leads to a quadratic in $m$.

Answer 3

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.