2
$\begingroup$

Find the values of $m$ in the 2nd degree equation $mx^2-2(m-1)x-m-1=0$ so that it has only one root between $-1$ and $2$.

Like in this almost identical question there are two ways to solve this, one is acknowledging that $f(-1)f(2) \lt 0$, the other applying the theorems for the two possible scenarios $x_1 \lt -1 \lt x_2 \lt 2$ and $-1 \lt x_1 \lt 2 \lt x_2$.

Doing $f(-1)f(2)<0$, we have: $ \begin{align*} \left(m(-1)^2-2(m-1)(-1)-m-1\right)\left(m2^2-2(m-1)2-m-1\right)&\lt0\\\ (2m-3)(-m+3) & \lt 0 \\\ -2m^2+9m-9 & \lt 0 \\\ m \lt \frac{3}{2} \quad \vee \quad m \gt 3 \end{align*} $

The thing is that this answer is wrong since the correct one is $m \lt \frac{3}{2} \quad \wedge \quad m \neq 0 \quad \vee \quad m \gt 3$. How do I get to know and not only verify that $m \neq 0$ what did I do wrong?

Using the theorems in the scenarios I.$ \quad x_1 \lt -1 \lt x_2 \lt 2$ and II.$ \quad -1 \lt x_1 \lt 2 \lt x_2$, we have:

I. $$x_1 \lt -1 \lt x_2 \lt 2 \implies \Bigl(af(-1) \lt 0\Bigr) \wedge \Bigl(af(2) \gt 0\Bigr) \wedge \Bigl(\Delta \gt 0 \Bigr)\wedge \Bigl( \frac{S}{2} \lt 2\Bigr).$$

II. $$ -1 \lt x_1 \lt 2 \lt x_2 \implies \Bigl( af(-1) \gt 0\Bigr)\wedge\Bigl( af(2) \lt 0\Bigr) \wedge \Bigl(\Delta \gt 0 \Bigr) \wedge\Bigl(\frac{S}{2} \gt -1 \Bigr)$$

So both I and II have the condition $\Delta \gt 0$ in common. \begin{align*} \Delta \gt 0 &\implies [-2(m-1)]^2-4m(-m-1) \gt 0\\\ &\implies 8m^2-4m+4 \gt 0 \\\ \end{align*}

The problem is that the $\Delta$ from the equation $8m^2-4m+4=0$ is negative. $\Delta = b^2-4ac = (-4)^2-4(8)(4) \lt 0 \implies \nexists m$ such that $8m^2-4m+4=0 \implies \nexists m$ such that $[-2(m-1)]^2-4m(-m-1) \gt 0$. So doesn't matter what the others conditions says in I (or in II), intersecting them all will give an empty set. And therefore I $\cup$ II will give $\emptyset$.

So I am doing it wrong both ways. Any thoughts on that?

  • 0
    Your formatting really makes things hard to read and follow. You don't put arrows on both previous and next line, just on one of them. And don't use `.` for multiplication! It's too much like a decimal point. Either use `\times` or `\cdot` (the latter will produced a raised dot)2011-01-03
  • 0
    I am going to correct it now. Thanks. Right, I'll use \cdot or \times next time.2011-01-03
  • 0
    I would like someone to address the theorems solutions, please.2011-01-03
  • 0
    There are no "theorems solutions" here. There is your solution to the problem, and I *did* address it. I explained what your error was: it lies in assuming that because the discriminant of $\Delta$ is never zero, this means that $\Delta$ itself is never positive. That is not the case.2011-01-03

2 Answers 2