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I guess you guys will enjoy this problem (taken from Brilliant.org) -

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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spit that fire
Find the sum of all integral values of \(\alpha \in [-20,20]\) such that\[2\log(x+3) = \log(\alpha x)\]has exactly one real solution.
So first things first:\[x> -3\tag1\]\[\alpha x > 0 \tag 2 \]If we accept these two conditions, we can now recast the problem into the form.\[\log\left((x + 3)^2\right) = \log(\alpha x)\]\[\Rightarrow (x+3)^2 = \alpha x\]\[\Rightarrow x^2 + (6 - \alpha)x + 9 = 0\]Pay attention to the fact that the quadratic can have one real solution, but that is not the only possibility. The quadratic may as well have two real solutions, yet one of them making the logarithm undefined. So let's first see when the quadratic itself has one real solution. I'll save us the work and tell you guys that it occurs at \(\alpha = 0, 12\) but \(\alpha = 0\) is eliminated because of the second condition of course. Now comes the hard part in my opinion.

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|dw:1434998964257:dw|
So let's try to check where \(x^2 + (6 - \alpha) x + 9\) has one real root lying in the interval \((\infty, -3]\). To have two real roots, \((6 - \alpha)^2 > 36\) o \(\alpha(\alpha - 12) > 0\) meaning \(\alpha \in (-\infty, 0) \cup (12, \infty)\). To have one root in the interval \((-\infty, -3)\) ...
we know that \(f(-3) < 0\). Thus \(9 + 9 -3(6 - \alpha) < 0 \Rightarrow \alpha < 0\).
Ooo, interesting.
That looks neat! for an alternative, im thinking of using calculus and mess with slopes the system has an unique solution when the slope of line is not so steep enough to cut the parabola at two different places and not too low to miss the parabola completely need to work integral values separately
But if \(\alpha < 0\), we also know that the other root must be in the interval \((-3,0)\) since \(\alpha x>0\).
Do we really need to solve for the above condition? We already know that the roots of the equation \(x^2 + (6 - \alpha) x + 9\) have the same sign so wouldn't the second condition be covered?
I'm not as experienced as you when it comes to calculus - but go right ahead.
OK, \(\alpha = -20, -19, \cdots, -1, 12\) so far.
hey pk :)
Where did Ganeshie leave?
I'm missing steef. :(
I think that's it - we can't really make the \(\log(\alpha x)\) undefined for one value and defined for the other since both roots have the same sign, right? So they're in this together.
Amazing, I got it right!
That isn't as complicated as I thought it is going to be!
:)

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