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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?

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  • $\begingroup$ I'm afraid it is not. Could you show your work, so that we can provide appropriate guidance if necessary? For instance, what are the expressions of $x_1$ and $x_2$ in terms of $m$? $\endgroup$ – Demosthene Jan 4 '15 at 0:51
  • $\begingroup$ The conditions are $\Delta>0$, $(2m+1)[(2m+1)2^2-(4m+2)2+m-1)]>0 $ and $(4m+2)/2(2m+1)<2$. The last is always true. So by solving the first two I obtain $(-\frac{1}{2};-\frac{1}{3})$. $\endgroup$ – chen h. Jan 4 '15 at 0:55
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This king of problem can be solved without knowing the exact values of the roots.

  • The equation must have real two roots; a 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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  • $\begingroup$ Your first condition is not enough because if 2m+1<0 then P(2)<0 also satisfies the condition for the roots. $\endgroup$ – chen h. Jan 4 '15 at 1:46
  • $\begingroup$ @chen h. You're right, I did as though the coefficient of $x^2$ was $1$. I'll fix that. Thanks for pointing this problem. $\endgroup$ – Bernard Jan 4 '15 at 1:48
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The discriminant is $$\Delta=b^2-4ac=(4m+2)^2-4(2m+1)(m+1)=8m^2+20m+8$$ To get real roots, $m$ must satisfy $$\Delta>0$$ which reduces as $$2m^2+5m+2>0$$ and has for solution $m>-1/2$ or $m<-2$. The roots are then $$x_{2,1}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{4m+2\pm\sqrt{8m^2+20m+8}}{4m+2}$$ with $x_1=1-\frac{\sqrt{8m^2+20m+8}}{4m+2}$ and $x_2=1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$.

If we want $x_1<x_2$, then $$1-\frac{\sqrt{8m^2+20m+8}}{4m+2}<1+\frac{\sqrt{8m^2+20m+8}}{4m+2}$$ which yields $m>-1/2$, consistent with what we found above.

Now, we also want $x_2<2$, so we solve $$\frac{\sqrt{8m^2+20m+8}}{4m+2}<1$$ and we find $m>1$ or $m\le-2$.

Combining all these conditions, if we want real roots $x_1$ and $x_2$ that verify $x_1<x_2<2$, we need $$m\in\left]-\infty;-2\right[\cup\left]1;+\infty\right[$$

Of course, you should always check these results, and also interpret them graphically. For instance, using Mathematica, the command

Manipulate[Plot[(2*m + 1)*x^2 - (4*m + 2)*x + m - 1, {x, -1, 3}], {m, -10, 10}]

creates a nice parametric plot and you can observe the behaviour of the solutions as they approach the boundaries.

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  • $\begingroup$ It would be better if you can divide the discussion into 2 cases, one for m>-1/2 and one for m<-2. This is because, when m<-2, -6>4m+2, a negative quantity. Then, your choice of $x_1=1-\frac{\sqrt{8m^2+20m+8}}{4m+2}$ will therefore be larger than $x_2$, violating the required condition. $\endgroup$ – Mick Jan 4 '15 at 5:07
  • $\begingroup$ @Mick Agreed. But the OP does not give precise definitions of $x_1$ and $x_2$ (to what specific root they correspond). You'd indeed have to reverse the order of $x_1$ and $x_2$ for $m<-2$; but the way the question is formulated leaves some room for ambiguity, and getting two distinct roots, both smaller than 2 is the way I read it. If the OP were to be more specific, I would adapt my answer. $\endgroup$ – Demosthene Jan 4 '15 at 14:24
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Such problems can be solved if you use geometrical interpretation (a quadratic expression representing parabola) and then use that to form the required inequalities.

For example, if we want to get the conditions for the roots $x_1,x_2$ of $ax^2+bx+c=0$ such that $$x_1 < x_2 < s \qquad \text{ for some fixed } s \in \mathbb{R}$$

(1). First condition: We want two real and unequal roots: for that we need $$b^2-4ac > 0 \qquad \text{ and } \qquad a \neq 0$$

(2). Second condition: We want the roots to be satisfy the given inequality. Then we need to consider two cases: parabola facing up or parabola facing down. Let $f(x)=ax^2+bx+c$.

  • Parabola facing up (and both roots lie to the left of $s$): For this we want the following conditions: $$a>0 \qquad \text{ and } \qquad f(s)>0.$$

    • Parabola facing down (and both roots lie to the left of $s$): For this we want the following conditions: $$a<0 \qquad \text{ and } \qquad f(s)<0.$$

The common solutions to all the inequalities from the first and second conditions will give you your required result.

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