Question

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

Answer 1

spit that fire

Answer 2

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.

Answer 3

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.

Answer 4

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 \((-\infty, -3)\) ...

Answer 5

we know that \(f(-3) < 0\). Thus \(9 + 9 -3(6 - \alpha) < 0 \Rightarrow \alpha < 0\).

Answer 6

Ooo, interesting.

Answer 7

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

Answer 8

But if \(\alpha < 0\), we also know that the other root must be in the interval \((-3,0)\) since \(\alpha x>0\).

Answer 9

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?

Answer 10

I'm not as experienced as you when it comes to calculus - but go right ahead.

Answer 11

OK, \(\alpha = -20, -19, \cdots, -1, 12\) so far.

Answer 12

@ikram002p Hey.

Answer 13

hey pk :)

Answer 14

Where did Ganeshie leave?

Answer 15

I'm missing steef. :(

Answer 16

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.

Answer 17

Amazing, I got it right!

Answer 18

That isn't as complicated as I thought it is going to be!

Answer 19

:)