ParthKohli
  • ParthKohli
I guess you guys will enjoy this problem (taken from Brilliant.org) -
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
spit that fire
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
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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ganeshie8
  • ganeshie8
|dw:1434998964257:dw|
ParthKohli
  • ParthKohli
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 \((-\infty, -3)\) ...
ParthKohli
  • ParthKohli
we know that \(f(-3) < 0\). Thus \(9 + 9 -3(6 - \alpha) < 0 \Rightarrow \alpha < 0\).
ParthKohli
  • ParthKohli
Ooo, interesting.
ganeshie8
  • ganeshie8
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
ParthKohli
  • ParthKohli
But if \(\alpha < 0\), we also know that the other root must be in the interval \((-3,0)\) since \(\alpha x>0\).
ParthKohli
  • ParthKohli
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?
ParthKohli
  • ParthKohli
I'm not as experienced as you when it comes to calculus - but go right ahead.
ParthKohli
  • ParthKohli
OK, \(\alpha = -20, -19, \cdots, -1, 12\) so far.
ParthKohli
  • ParthKohli
@ikram002p Hey.
ikram002p
  • ikram002p
hey pk :)
ParthKohli
  • ParthKohli
Where did Ganeshie leave?
ParthKohli
  • ParthKohli
I'm missing steef. :(
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
Amazing, I got it right!
ganeshie8
  • ganeshie8
That isn't as complicated as I thought it is going to be!
ikram002p
  • ikram002p
:)

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