For which values of $m$ the equation
$$(2m+1)x^2-(4m+2)x+m-1=0$$
has two real roots $x_1$ and $x_2$ that satisfy the condition: $x_1<x_2<2$?
I found that $$m\in\left(-\frac{1}{2}; -\frac{1}{3}\right).$$ Is it correct?
algebra-precalculus
For which values of $m$ the equation
$$(2m+1)x^2-(4m+2)x+m-1=0$$
has two real roots $x_1$ and $x_2$ that satisfy the condition: $x_1<x_2<2$?
I found that $$m\in\left(-\frac{1}{2}; -\frac{1}{3}\right).$$ Is it correct?
Best Answer
This king of problem can be solved without knowing the exact values of the roots.
reduced discriminant
$\Delta'=(2m+1)^2-(2m+1)(m-1)=(2m+1)(m+2)$.$$ \Delta'>0\iff m<-2 \enspace \text{or}\enspace m>-\dfrac 1 2$$
Note. Since the equation has the form: $ax^2+2b'x+c$, we used $\Delta'=b'^2-ac$ instead of $\Delta$.
The formulae for the roots then simplify to: $\enspace x_1,x_2=\dfrac{-b'\pm\sqrt{\Delta'}}{a}$
$2$ must
not
separate the roots. This means that $aP(2)>0$, i.e. $(2m+1)(m-1)>0$. Taking into account the condition on $\Delta'$, this means $m>1$ or $m<-2$. Then $2$ is either greater than or smaller thanboth
roots.Knowing that, $2$ is greater than the roots if and only it is greater than their
arithmetic mean
. This mean can be expressed with the coefficients: $x_1+x_2=-\dfrac b a $, so here we have: $\dfrac{x_1+x_2}{2}=1$. Indeed $\enspace2>\dfrac{x_1+x_2}{2}$.Finally, we know $x_1<x_2<2$ if and only if $ m >1$ or $m<-2$.