# For which values of a parameter an equation has one Real root

The following equation is given $$\log_{x-1}(x^2+2ax) - \log_{x-1}(8x-6a-3)=0$$

And I am trying to find for which values of $a$ it has only one root, which is real.

It is obvious that $$x-1>0 \Rightarrow x>1, x\neq2$$

Also $$x^2+2ax>0$$ $$8x-6a-3>0 \Rightarrow x>\frac{6a+3}8$$

Now I combine the two logarithms $$\log_{x-1}\left(\frac{x(x+2a)}{8x-6a-3}\right)=0$$ and then $$\frac{x(x+2a)}{8x-6a-3}=1$$ $$\frac{x^2+2(a-4)x+6a+3}{8x-6a-3}=0$$ $$x^2+2(a-4)x+6a+3=0$$ The discriminant is $$D=(a-13)(a-1)$$ and the roots are $$x_1= -a+4-\sqrt{(a-13)(a-1)}$$ $$x_2= -a+4+\sqrt{(a-13)(a-1)}$$ so that $x_1<x_2$.

• Case 1:

For $D=0$ we have only one root for $x$. $D=0$ for $a=13$ or $a=1$.

For $a=13$, $x_1=x_2=-9\not\gt1$ so $a=13$ is not a solution.

For $a=1$, $x_1=x_2=3>1$. Also $x=3>\frac{6a+3}{8}=\frac98$ and $x^2+2ax=15>0$, so for $a=1$ we have only one root, therefore it is a solution.

• Case 2:

Now let $D>0$, so that $x_1 \ne x_2$. $D>0$ for $a\in(-\infty,1)\cup(13,+\infty)$

I consider five sub-cases:

A) When $x_1 = 2$ and $x_2^2+2ax_2>0, x_2>\frac{6a+3}8$

B) When $x_2=2$ and $x_1>1, x_1^2+2ax_1>0, x_1>\frac{6a+3}8$.

C) When $1\in[x_1,x_2)$ and $x_2^2+2ax_2>0, x_2>\frac{6a+3}8$.

D) When $\frac{6a+3}8\in[x_1,x_2)$ and $x_2>1, x_2^2+2ax_2>0$.

E) When $x_1^2+2ax_1\le0$ and $x_2>1, x_2^2+2ax_2>0, x_2>\frac{6a+3}8$

The solutions I find are

A) $a=\frac9{10}, x_2=\frac{21}5$

B) none

C), D), E) It all comes up to solving inequalities involving polynomials for $a$ of degree 4, which makes me doubt I am doing it right.

Am I missing any cases or am I considering too many cases? How would you approach this problem?

• you can divide $x_1,x_2$, so you will have a quadratic inequality to solve. – chenbai Jun 4 '15 at 0:09
• if you use Vita's theory, it will much easy. – chenbai Jun 4 '15 at 4:37

For $D=0$, $a=1$ is the only solution.

For $D=(a-1)(a-13)\gt 0$, note that $x_2\not=2$ and that $x_1=2$ for $a=\frac{9}{10}$ (and $a=\frac{9}{10}$ is sufficient). In the following, let us separate it into cases.

• For $a\lt -\frac 12$, we have $\frac{6a+3}{8}\lt 0\lt 1\lt -2a.$ So, the condition we need is $x_2\gt -2a$ and $x_1\le -2a$, i.e. $$\small a\lt -\frac 12\ \ \text{and}\ \ -a+4+\sqrt{(x-1)(x-13)}\gt -2a\ \ \text{and}\ \ -a+4-\sqrt{(a-1)(a-13)}\le -2a.$$Solving these gives you $a\lt -\frac 12$.

• For $-\frac 12\le a\lt 0$, we have $0\lt -2a\le 1$ and $0\le \frac{6a+3}{8}\lt 1$. So, the condition we need is $x_2\gt 1$ and $x_1\le 1$, i.e. $-\frac 12\le a\lt 0$.

• For $0\le a\lt\frac 56$, we have $-2a\le 0\lt\frac{6a+3}{8}\lt 1$. So, the condition we need is $x_2\gt 1$ and $x_1\le 1$, i.e. $0\le a\le \frac 12$.

• For $\frac 56\le a\lt \frac{9}{10}\ \text{or}\ \frac{9}{10}\lt a\lt 1\ \text{or}\ a\gt 13$, we have $-2a\lt 0\lt 1\le\frac{6a+3}{8}$. So, the condition we need is $x_2\gt\frac{6a+3}{8}$ and $x_1\le\frac{6a+3}{8}$. However, there is no such $a$.

Hence, the answer is $$a\le\frac 12\ \ \ \text{or}\ \ \ a=\frac{9}{10}\ \ \ \text{or}\ \ \ a=1.$$