Find all $m \in R$ such that $(m + 1)x^2 -2(m - 3)x + m + 1 = 0$ has at least one root such that its absolute value is smaller than $0.9$ Find all $m \in R$ such that equation $(m + 1)x^2 -2(m - 3)x + m + 1 = 0$ has at least one root such that its absolute value is smaller than $0.9$
If $m = 1$ this equation has only one solution $x = -1$ which does not meet the requirements. If $m = -1$ then $x = 0$, which meets the requirements. Now I can assume that $m < 1$ and the equation has two solutions $x_1$ and $x_2$. 
Is there any way to solve it faster than just substituting $x_1 = \frac{-b  +/- \Delta}{2a}$ into $|x_1| < 0.9$ and doing the same thing for $x_2$?
 A: You can also try working in reverse, and work out the equation for $m$ (which is linear in $m$).  The equation for $x\neq 1$ is $$m=\frac {8-(x+3)^2}{(x-1)^2}$$
What are the values of $m$ when $-0.9\lt x\lt 0.9$?
A: You can calculate the value of $(m+1)x^2−2(m−3)x+m+1$ at $x=\pm0.9$. If the product of these values is negative, you have exactly one solution in this interval. If the product is positive, check the value at one end, and the value for $x=-\frac{b}{2a}$. If they have opposite signs, both solutions are in the interval. If they have the same sign, there is no solution in $[-0.9,0.9]$
A: You can use the following result from high school:

A number $\xi\in\mathbf R$ separates (in the extended sense: $\xi$ can be one of the roots, and these can be equal) the (real) roots of a quadratic polynomial $p(x)=ax^2+bx+c$ if and only if
  $$ \mathrm{(i)}\; b^2-4ac\ge0,\qquad\mathrm{(ii)}\;ap(\xi)\le 0. $$

Thus we first have to check whether there are real roots. The condition is the reduced discriminant
$$\Delta'=(m-3)^2-(m-1)^2=-4(m-2)\ge 0\iff m\le 2.$$
Next (for $m\le2$), there are three cases: either


*

*$-0.9$ separates the roots, but $0.9$ does not. This translates into
\begin{cases}(m+1)p(-0.9)=(m+1)\bigl[1.81(m+1)+1.8(m-3)\bigr]\le 0\\
\phantom-(m+1)p(0.9)=(m+1)\bigl[1.81(m+1)-1.8(m-3)\bigr]> 0
\end{cases}

*$0.9$ separates the roots, but $-0.9$ does not. This translates into
\begin{cases}(m+1)\bigl[1.81(m+1)+1.8(m-3)\bigr]> 0\\
(m+1)\bigl[1.81(m+1)-1.8(m-3)\bigr]\le 0
\end{cases}
Having one of theses cases is equivalent to 
\begin{align}
(m+1)^2p(0.9)p(-0.9)\le 0&\iff p(0.9)p(-0.9)\le 0\iff 1.81^2(m+1)^2-1.8^2(m+3)^2\le 0\\
\iff(0.01m-3.61)(3.61m+7.21)\le 0&\iff \frac{721}{361}\le m\; (\le 2)\le 361
\end{align}

*Neither separates the roots, but the  mean of the roots $\dfrac{m-3}{m+1}$ separates $-0.9$ and $0.9$, i.e. taking into account that $m\le 2$:
$$\Biggl|\frac{m-3}{m+1}\Biggr|=\frac{3-m}{|m+1|}\le 0.9\iff3-m\le 0.9\,|m+1|\iff\begin{cases} m\ge\strut\smash{\dfrac{21}{19}}&\text{if}\;m>-1\\[1.5ex]m\ge 39&\text{if}\;m<-1
\end{cases}$$


The last case can't happen, so the end result is
$$\frac{21}{19}\le m\le 2.$$
