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So I was doing a math exam and I came across a question and while trying to figure it out I came across what looked like proof for 0/0 and while I know it's probably not I was hoping that someone could explain why as it confused me. Most of the details for the question are irrelevant but I'll provide them for context. The question was along the lines of:

There is a 9m tall arch that follows the shape of a parabola and the base of the arch is 6m. A builder wants to know how long a plank of wood must be to span the arch at 4m up.

Basically to solve it I put the graph on a cartesian plane with the turning point on the y-axis and used the turning point form of a parabola: y=ax^2+9

To solve for a, I subbed the turning point into the equation and rearranged, and this is where I came along what looked to be proof for 0/0:

9=a0^2+9 [-9] 0=a0 [/0] 0/0=a

And when I got to that I realised that that wouldn't work so I instead used one of the x-intercepts and found that a=-1:

0=a3^2+9 [-9] -9=a9 [/9] ∴ a=-1

Now when I look back on it I realise that a could have be any number and I still would have run across this issue and I'm not sure why this is. I'm thinking it may be why 0/0 is undefined as it could represent any number? If someone could explain this to me and help it would be appreciated.

Thanks!!

Intelligenti pauca
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1 Answers1

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Think of it this way:

The base of the parabola is 6m.

So, one tip of the parabola on the ground is at +3m and the other at -3m

Now, we can solve for $a$: Write $y=ax^2+9$ and fill in $y=0$ and $x=3$. We get $a=-1$, as you got.

The problem at $x=0$ is that you try to solve for $a$, after $a$ should be eliminated from the equation: Filling in $x=0$ into $y=ax^2+9$ gives $y=9$ and you cannot solve for $a$. Said otherwise, whatever you choose for $a$, we always have $y=9$ at $x=0$.

Said otherwise again, indeed $0/0$ is "undefined" because it can be any number. Saying $x=0/0$ means that $x$ solves the equation $0x=0$ (similar to how $x=4/2$ solves the equation $2x=4$). Every $x$ solves that equation! See also e.g. this question

student91
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