The answer is not 12

Algebra Level 5

Find the sum of integral values of $\alpha$ such that $2\log (x+3)=\log (\alpha x)$ has exactly one real solution and $|\alpha |<20$ .

The answer is -178.

2 solutions

Pranjal Jain
Apr 16, 2015

$\log (x+3)^2=\log(\alpha x)\Rightarrow x^2+(6-\alpha)x+9=0$

Now, this quadratic must have exactly one solution in the domain of given equation.

Case 1: Determinant is $0$ and the value of $x$ is in the domain,

$(6-\alpha)^2=36\Rightarrow \alpha=0, 12$

Due to domain constraint, $\alpha\ne 0$ , $\alpha=12$ gives $x=3$ which satisfies the domain.

Case 2: Determinant is positive but one of the root is in domain,

Case 2.1: $\alpha>0$ , in this case $x$ has to be positive. So the quadratic equation must have a positive and negative root, which is not possible as the product of roots is positive (ie, 9)

Case 2.2: $\alpha<0$ , in this case, $x$ has to be negative. So the quadratic equation must have a root in $(-3,0)$ . By using location of roots, all negative values of $\alpha$ satisfies this.

Thus, $\alpha\in (-\infty,0)\cup \{12\}$

Sum of all possible values is $12+(-1)+(-2)+\cdots+(-19)=12-\dfrac{19\times 20}{2}=12-190=\boxed{-178}$

Ahh .. I did wrong calculation ... so i lost my chances ... But Thanks for Posting this good Problem . Please Post more

Here's Solution...

$\displaystyle{{ D }_{ f }:\quad x\in (-3,0)\cup (0,\infty )\\ case(i):\quad -3<x<0\\ \therefore \alpha <0\quad ...........(A)\\ f\left( x \right) ={ x }^{ 2 }-(\alpha -6)x+9=0\\ \Delta =\alpha (\alpha -12)\quad \\ \alpha <0\quad \Rightarrow \Delta >0\\ \because f\left( 0 \right) =9>0\\ \therefore f(-3)<0\quad ...(unique\quad root)\\ \Rightarrow \alpha <0\quad .......(B)\\ A\cap B\\ \boxed { \alpha <0 } \quad ......\quad (I)\\ case(ii):\quad 0<x<\infty \\ \therefore \alpha >0\quad .........(C)\\ \Delta =0\quad \Rightarrow \quad \alpha =0,12\quad .......(D)\\ C\cap D\quad \\ \boxed { \alpha =12 } \quad .......\quad (II)\\ I\cup II\\ \boxed { \alpha \in { R }^{ - }\cup \left\{ 12 \right\} } \\ }$

Karan Shekhawat - 6 years, 1 month ago

@Pranjal Jain You have a fan now. Wow.

Kunal Verma - 6 years, 1 month ago

Thanks! Haha!

Pranjal Jain - 6 years, 1 month ago

Nice solution. I started out much the same way, but then decided that it is much easier if you consider $\alpha$ as a function of $x$ (only defined for $x>-3$ of course).

Peter Byers - 6 years, 1 month ago

@Calvin Lin Sorry, I was leaving for the class. I hope its fine now. You may remove the note.

Pranjal Jain - 6 years, 1 month ago

Thanks! Yes, this now explains why the solution includes the negative (real) numbers )

Calvin Lin Staff - 6 years, 1 month ago

I think you should perhaps state that $x$ is real.

Jake Lai - 6 years, 1 month ago

Indeed. Thanks.

Pranjal Jain - 6 years, 1 month ago

for the negative values of alpha does x have EXACTLY one solution..i mean cant the parabola have two olutions between 0 and -3 ??

Writo Mukherjee - 5 years, 11 months ago

@Writo Mukherjee – Nope, as one of the roots will be positive

Pranjal Jain - 5 years, 11 months ago

Why is the ans -178 ,explain!

Ankit Bajaj - 6 years, 1 month ago

Added the last part. You may check it now.

Pranjal Jain - 6 years, 1 month ago

why in case 2.1 for positive x we have +ve and -ve root .explain

Ankit Bajaj - 6 years ago

Checking the domain, $\alpha x>0$ . Since $\alpha>0$ , we need $1$ positive $x$ and the other $x$ negative. So at the end, we'll neglect negative one. Thus, getting only one solution, the positive one.

Pranjal Jain - 6 years ago

Critical question!!

Arunava Das - 3 years, 5 months ago

I think yo mean discriminant instead of determinant :). I also mix them up so easily XD

Nicolas Nauli - 2 years, 1 month ago

Well, all my answer is wrong hahahaha. By the way, this is beautiful problem. :D

Paul Ryan Longhas - 6 years, 1 month ago

Carlos Victor
Apr 23, 2015

Observing the graphs of f (x) = (x + 3) ^ 2 and g (x) = kx, with x> -3 and x non-zero; we find that for k = 12 (g is tangent to the graph of f) and -20 <k <0 we have a unique solution of f(x)=g(x). We will soon have to sum k : -1-2-3 ...- 19 +12 = -178.

Hi Pranjal Jain : Looking at the intersections of the graphs of f and g functions mentioned above, we note the following: for x> -3, x <0 and k <0, there is only one intersection, and since k is an integer and | k | <20, the possible values are : -19, -18, ...- 1.

Looking at the graphs f and g for x> 0 and k> 12, have two intersections. For k = 12 we will have a single intersection and 0 <k <12 will not have intersection.

How'd you know that you have to check these k's only? Or that there are no more such k's?

Pranjal Jain - 6 years, 1 month ago

Hi Pranjal: Please, look at the above comment in the question itself. hugs

Carlos victor - 6 years, 1 month ago

The answer is not 12

The answer is -178.

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