For what values of $m$ ​does $x^2+2\sqrt3x-\log m = 0$ admit real roots?

Question

For what values of $m$ ​​does the equation $$x^2+2\sqrt3x-\log m = 0$$ admit real roots? What are the signs of the roots of the equation for these values ​​of $m$?

I try

$\Delta \geq 0 \implies 12+4\log m \geq 0 \implies \log m \geq -3\\ \therefore m \geq \frac{1}{10^3}$

Answer 1

By solving for $x$ as a simple quadratic, we find

$$x = \pm \sqrt{3 + \log m} - \sqrt 3$$

$x$ is real if and only if $\log m \ge -3$, and therefore $x$ is real whenever $m \ge 1/e^3$.

The only difference between my answer and yours is that I'm assuming $\log$ to mean $\ln$. So basically you are correct.

One root will always be negative since $-(\sqrt \cdot + \sqrt3)$ is always less than zero, since $\sqrt \cdot$ is always at least zero. As for the sign of the other root.. Simplifying $\sqrt{3 + \log m} - \sqrt 3 \ge 0$ by adding $\sqrt3$, squaring both sides, and cancelling the remaining 3's, we find that the other root is non-negative if and only if $\log m \ge 0$, that is, that $m \ge 1$.

• $m = 1/e^3 \implies -^2$, that is one negative root of multiplicity 2.
• $ 1/e^3 < m < 1 \implies --$, that is two distinct negative roots.
• $ m = 1 \implies -0$, that is one negative, and a root at $x=0$.
• $ m > 1 \implies -+$, that is one negative and one positive root.

As a side note, the reason that $x > y$ if and only if $\log x > \log y$ if and only if $e^x > e^y$ is that $\exp$ and its inverse $\log$ are order-preserving. The index of the maximum of $a, b, c, \cdots$ will be the same as $\log a, \log b, \log c, \cdots$ will be the same as $\exp a, \exp b, \exp c, \cdots$. The map $x \mapsto 1/x$ is the one you need to be careful about with inequalities, but I won't discuss it here since it isn't relevant to the problem. Just sharing a bit of inequality wisdom here seemed appropriate.

Answer 2

Hint: If $x_1$ and $x_2$ are the roots of the quadratic equation $x^2+px+q=0$, then the following equalities hold:

$$x_1+x_2=-p, \ \ \ \ \ \ x_1\cdot x_2=q.$$

Answer 3

Completing the square: $$x^2+2\sqrt3 x -\log m=0$$ $$x^2+2\sqrt3 x +3=3+\log m$$ $$(x+\sqrt3)^2=3+\log m$$ Now, if $$3+\log m\geq 0$$ then we have real roots. So, $$\log m\geq -3$$ and this means $$m\geq 10^{-3}.$$

The roots are $x_{1,2}=\pm\sqrt{3+\log m}-\sqrt3.$

The signs of the roots: $$ \begin{array}{c|c} m & \text{Signs} \\ \hline m=10^{-3} & -^{*}\\ 10^{-3}<m<1 & -,- \\ m=1 & -,0^{**} \\ 1<m&-,+ \end{array} $$

$*:$ There is one root of multiplicity two which is $x=-\sqrt3$.

$**:$ The roots are $x_1=-2\sqrt3, x_2=0.$

Answer 4

One way to attack your question with a perceptive yet deeper objective, is to understand what the "$-\log{m}$" term actually does to the parabola.

In its expanded form, $ax^2 + bx + c$, observe the $c$ term has no $x$ attached and hence only translate $ax^2 + bx$ in a rectilinear way. Specifically, all ordinates of $ax^2 + bx$ are simply offset by $c$. For a concave parabola, its vertex is the minimum value. If $c = 0$, then it is given by $f\left(\dfrac{-b}{2a}\right) = \dfrac{-b^2}{4a}$. For your quadratic, treating $-\log{m} = 0$, it would be $\left(-\sqrt{3}, \dfrac{-(2\sqrt{3})^2}{4} = -3 \right)$.

See that $-\log{m} = 0$ meets requirements of your question to still admit real roots since the vertex is below $y = 0$. Now ask yourself, how can $-3$ be offset, by $c = -\log{m} \neq 0$ such that it still admits real roots. Now one can clearly see why $c = -\log{m} = 3$ is the minimum, as thats when the vertex is allowed to touch the x-axis. $\therefore \log{m} > -3$.

Now, lets observe $-\log{m}$ affects the signs of the roots. Assume, $c = -\log{m} = 0$ and the two roots be $x_1 \le x_2$. Then we can factor to $x(ax + b) = 0$, denoting that $x_2 = 0$ is always a root when $c = 0$. The other root is $x_1 = \dfrac{-b}{a} = -2\sqrt{3}$. Note that $x$ coordinate of the your vertex, will always be $-\sqrt{3}$, since its not affected by $c$. Hence, root on its left, $x_1 = 2\sqrt{3}$ will always be $-$, no matter $\log{m}$.

Now, observe if $\log{m} > 0$, the parabola will move in the negative direction vertically (downwards) and $f(x = 0) < 0$. This clearly means the $x_2$ must hit in the $+x$.

Conversely, if $-3 \le \log{m} < 0$, will move in the positive direction vertically (upwards) and $f(x = 0) > 0$. This means the $x_2$ must hit in the $-x$.

The case of $\log{m} = 0, x = 0$ should be accepted too; $x_2$ will be neutral.

Hence, the all cases simplified are $\displaystyle\begin{cases}{\{m = 10^{-3}; \space (x_1, x_2) = (-^2)\}\\ \{10^{-3} < m < 1; \space (x_1, x_2) = (-, -)\} \\ \{m = 1; \space (x_1, x_2) = (-, 0)\}} \\ \{m > 1; \space (x_1, x_2) = (-, +)\}\end{cases}$

I understand this maybe a weighty explanation but once you grasp the concept, the mathematical algebra is effortless to write (as seen by the other answers) and the concept can be quickly employed within the thinking process itself.