Determine $m \in \mathbb{R}$ so that the inequality is true for all $x \in \mathbb{R}$

Question

What is the method to go about to solve it for:

$mx²+(m-1)x+(m-1) < 0$

Sorry if this seems too easy for the website standard, but I am working on this alone, and couldn't find a way on my own.

Answer 1

To find values of $m$ for which $mx^2+(m-1)x+(m-1)$ is negative for all $x\in\mathbb R$, first note that if

$m\ge0$ then $mx^2+(m-1)x+(m-1)$ takes arbitrarily large positive values, so we need $m<0$.

For $m<0$, the maximum of $mx^2+(m-1)x+(m-1)=ax^2+bx+c$ is

$c-\dfrac {b^2}{4a}=(m-1)-\dfrac{(m-1)^2}{4m}=\dfrac{(m-1)(4m-(m-1))}{4m}=\dfrac{(m-1)(3m+1)}{4m}$,

which is $0$ when $3m+1=0$, i.e., $m=-\dfrac13$; positive when $-\dfrac13<m<0$ (since then $m-1$ is negative, $4m$ is negative, and $3m+1$ is positive); and negative when $m<-\dfrac13$ (since then $m-1$, $3m+1$, and $4m$ are negative).

So the answer is that {$m\in \mathbb R|m<-\dfrac13$} is the subset of $\mathbb R$ for which $mx^2+(m-1)x+(m-1)$ takes on negative values for all $x \in \mathbb R$.