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Given a set D = {a+b•| a,b ∈ $\mathbb{R}$}

And a made-up binary operation on D is defined as follows: (a+b•)(c+d•)= ac+(ad+bc)•

For example, (2+3•)(-3+5•)= (-6+1•) You are not allowed to combine (-6+1•) into -5• because they are not like terms. you are allowed to combine like terms, however, like this: (a+b•)+(c+d•) = a+c+(b+d•)

So the question is: Solve the quadratic equation $x^2$-2x+12•=0

I'm very confused about the 12• and binary operation part. Should the quadratic formula be used here? How would you solve it?

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What, and why, is that dot to the right of some numbers? – DonAntonio Dec 1 '13 at 20:35
the dot doesn't represent multiplication or anything, it's just a symbol to differentiate like terms. it's similar to how 2q is different from 2 so you can't combine 2q+2, but q doesn't represent any value. So in this case the • is used to create the binary operation – gticecream8 Dec 1 '13 at 20:37
@Berci sorry about that, I fixed it. It's just an example – gticecream8 Dec 1 '13 at 20:55
up vote 1 down vote accepted

Let's denote the $\bullet$ rather by $q$, even if that doesn't represent any real number as value. So that, $q:=0+1\!\bullet$.

Now we have $q^2=q\cdot q=0+0\bullet=0$, and basically that implies the whole multiplication (just the same way as $i^2=-1$ and linearity generates the multiplication for complex numbers).

We have to solve $x^2-2x+12q=0$. Write up $x$ as $x=a+bq$ then we have $x^2=a^2+2abq$, so what is needed is: $$a^2-2a+(2ab-2b)q=-12q$$ Looking at the 'coordinates' on both sides, we need $a^2-2a=0$ and $2b(a-1)=-12$.

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am I allowed to solve those two "coordinates" to find a = 0 and plug a into the second equation to find b=6 ? (excluding all the q's because the q's are confusing to solve with and I can just add them in later.) I'm confused as to what x would end up being, since it represents (a+b). As in, what would the final answer be for x – gticecream8 Dec 1 '13 at 21:07
Yes, either $a=0$ or $a=2$ and the corresponding $b$ values are $6$ and $-6$, meaning two solutions: $x=6\bullet$ and $x=2-6\bullet$. (For example, for $x=6\bullet$, its square is $0$, so $x^2-2x=-2x=-12\bullet$ indeed.) – Berci Dec 1 '13 at 21:17
thank you so, so much – gticecream8 Dec 1 '13 at 21:21
a similar question that relates to the system but not to the quadratic equation. Is it possible to use the created binary operation to get an answer of (5+0q) where q stands in for ∙ ? I've been trying for awhile but 5 seems prime in this system, though I know it is definitely not. – gticecream8 Dec 2 '13 at 4:57

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