Fun with inequalities-1

Let $\alpha < \beta$ be the roots of $x^2-2x-a^2+1 = 0$ and $\gamma < \delta$ those of $x^2-2(a+1)x+a(a-1)=0$ . Then we have $\gamma < \alpha < \beta < \delta$ .

$\begin{cases} \alpha = \dfrac {2-\sqrt{4+4a^2-4}}{2} = 1-a \\ \beta = 1+a\end{cases}$

$\begin{cases} \gamma = \dfrac {2(a+1)-\sqrt{4(a+1)^2-4a(a-1)}}{2} = 1+a - \sqrt{3a+1} \\ \delta = 1+a + \sqrt{3a+1} \end{cases}$

$\begin{cases} 1+a - \sqrt{3a+1} < 1-a & \Rightarrow 2a < \sqrt{3a+1} \\ & \quad 4a^2-3a-1 < 0 \\ & \quad (4a+1)(a-1) < 0 \\ & \quad x \in \left(-\frac{1}{4}, 1 \right) \\ 1+a + \sqrt{3a+1} > 1+ a & \Rightarrow \sqrt{3a+1} > 0 \\ & \quad a \in \left(-\infty, -\frac{1}{3} \right) \end{cases}$

Therefore, the answer is $\boxed{x \in \left(-\frac{1}{4}, 1 \right)}$

This is my solution:

$x^2-2x-a^2+1=0$ ...1

$x^2-2(a+1)x+a(a-1)=0$ ...2

Doing a process called completing the square for equation 1 :

$(x-1)^2-a^2=0$

Solving for x:

$x=a \pm 1$

Completing the square for 2:

$(x-(a+1))^2 - 3a - 1 = 0$

Now subbing $x= a\pm 1$ into $(x-(a+1))^2 - 3a - 1$ :

$(1 \pm 1-a-1)^2-3a-1$

This has to be $< 0$ because of the nature of a parabola. **

$1 \pm 1-a-1)^2-3a-1 <0$

$a^2(\pm 1 -1)^2 -3a-1<0$

there are 2 conditions to be met:

$a^2(1 -1)^2 -3a-1<0$

$-3a<1$

$\boxed {a>-1/3}$

$a^2(-1 -1)^2 -3a-1<0$

$4a^2 -3a-1<0$

$(a-1)(4a+1) <0$

$\boxed {-1/4 <a<1}$

Therefore : $\boxed {-1/4 <a<1}$

**(note: If you are interested on why it is <0 draw a parabola opening upwards that has 2 roots choose a point between them and evaluate the parabola at that point)