[Math] Quadratic equation with parameter and conditions on the roots

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.

  • The equation must have real two roots; as the coefficient of $x$ is divisible by $2$, we use the 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 than both 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 $m<-2$.

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