Find the values ​of $k$ that satisfy $x_1> 1$ and $x_2 <1.$

Question

Here I found a question that caught my attention. But then I did not see it again. The question was:

The quadratic equation is given as follows:

$$(2k+1)x^2-kx+k-2=0$$

Find the values ​​of $k$ that satisfy $x_1> 1$ and $x_2 <1.$

My way:

$$x_1=\frac{k+\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)} \\ x_2=\frac{k-\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}\\ \begin{cases} x_1>1 & \\ x_2<1 & \end{cases} \Rightarrow \begin{cases} \frac{k+\sqrt{k^2-4(2k+1)(k-2)}-2(2k+1)}{2(2k+1)} >0 & \\ \frac{k-\sqrt{k^2-4(2k+1)(k-2)}-2(2k+1)}{2(2k+1)}<0 & \end{cases} \Rightarrow \begin{cases} 2(2k+1)\left(\sqrt{-7k^2+12k+8}-(3k+2) \right)>0 & \\2(2k+1)\left(\sqrt{-7k^2+12k+8}+(3k+2) \right)>0 & \end{cases} \Rightarrow 4(2k+1)^2×\left( -7k^2+12k+8-(3k+2)^2\right)>0 \Rightarrow -16k^2+4>0 \Rightarrow (2k-1)(2k+1)<0 \Rightarrow k\in \left(-\frac12 ; \frac 12 \right)$$

Actually, I doubt my solution. Can you confirm that the solution is right or wrong?

Thank you!

Answer 1

As I've already commented, it seems that you showed that if $x_1\gt 1$ and $x_2\lt 1$, then $k\in\left(-\frac 12,\frac 12\right)$, and that you have not shown that if $k\in\left(-\frac 12,\frac 12\right)$, then $x_1\gt 1$ and $x_2\lt 1$.

This answer shows in your method that $x_1\gt 1$ and $x_2\lt 1$ is equivalent to $k\in\left(-\frac 12,\frac 12\right)$.

$$\small\begin{align}&x_1\gt 1\quad \text{and}\quad x_2\lt 1\\\\&\iff 2k+1\not=0\quad\text{and}\quad (-k)^2-4(2k+1)(k-2)\gt 0\\\\&\qquad\qquad\text{and}\quad \frac{k-\sqrt{(-k)^2-4(2k+1)(k-2)}}{2(2k+1)}\lt 1\quad\text{and}\quad \frac{k+\sqrt{(-k)^2-4(2k+1)(k-2)}}{2(2k+1)}\gt 1 \\\\&\iff k\not=-\frac 12\quad\text{and}\quad \frac{6-2\sqrt{23}}{7}\lt k\lt \frac{6+2\sqrt{23}}{7}\\\\&\qquad\qquad \text{and}\quad \frac{-3k-2-\sqrt{-7k^2+12k+8}}{2k+1}\lt 0\quad\text{and}\quad \frac{-3k-2+\sqrt{-7k^2+12k+8}}{2k+1}\gt 0 \\\\&\iff k\in\left(\frac{6-2\sqrt{23}}{7},-\frac 12\right)\cup\left(-\frac 12,\frac{6+2\sqrt{23}}{7}\right)\\\\&\qquad\qquad \text{and}\quad \frac{-3k-2-\sqrt{-7k^2+12k+8}}{2k+1}\lt 0\quad\text{and}\quad \frac{-3k-2+\sqrt{-7k^2+12k+8}}{2k+1}\gt 0\\\\&\iff \left[k\in\left(\frac{6-2\sqrt{23}}{7},-\frac 12\right)\ \ \text{and}\ \ \sqrt{-7k^2+12k+8}\lt -3k-2\ \ \text{and}\ \ \sqrt{-7k^2+12k+8}\lt 3k+2\right]\\\\&\qquad\qquad\ \text{or}\quad \left[k\in \left(-\frac 12,\frac{6+2\sqrt{23}}{7}\right)\ \ \text{and}\ \ \sqrt{-7k^2+12k+8}\gt -3k-2\ \ \text{and}\ \ \sqrt{-7k^2+12k+8}\gt 3k+2\right]\\\\&\iff k\in \left(-\frac 12,\frac{6+2\sqrt{23}}{7}\right)\quad\text{and}\quad \sqrt{-7k^2+12k+8}\gt |3k+2|\\\\&\iff k\in \left(-\frac 12,\frac{6+2\sqrt{23}}{7}\right)\quad\text{and}\quad -7k^2+12k+8\gt (3k+2)^2\\\\&\iff k\in \left(-\frac 12,\frac{6+2\sqrt{23}}{7}\right)\quad\text{and}\quad k\in\left(-\frac 12,\frac 12\right)\\\\&\iff k\in\left(-\frac 12,\frac 12\right)\qquad\blacksquare\end{align}$$

Answer 2

To solve

$$x_1=\frac{k+\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}>1$$

you need to set for the existence of real solutions

$$k^2-4(2k+1)(k-2)\geq0$$

and distinguish two cases

$$2(2k+1)>0\implies \sqrt{k^2-4(2k+1)(k-2)}>2(2k+1)-k$$

$$2(2k+1)<0\implies \sqrt{k^2-4(2k+1)(k-2)}<2(2k+1)-k$$

Similarly for

$$x_2=\frac{k-\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}<1$$

Answer 3

It is OK. I would do it this way. Our condition is $(1-x_1)(1-x_2)<0$ so we are interested when is $g(1)<0$ where $g(x) = (x-x_1)(x-x_2)$. Since $g(x) = {f(x)\over 2k+1}$, where $f(x)=(2k+1)x^2-kx+k-2$ we have $$ g(1) = {f(1)\over 2k+1} = {2k-1\over 2k+1}<0$$ and this is when $k\in (-{1\over 2},{1\over 2})$.