Question

If $\alpha ,\beta$ are the roots of the equation $x^2-2x-a^2+1=0$ and $\gamma ,\delta$ are the roots of the equation $x^2-2(a+1)x+a(a-1)=0$, such that $\alpha ,\beta ϵ(\gamma ,\delta )$, then find the value of 'a'.

Answer 1

As $\alpha$ and $\beta$ are roots of $x^2-2x-a^2+1=0$,

$\alpha ,\beta =\frac{2\pm \sqrt{4-4(1+a^2)}}{2}=1\pm a$

As $\gamma$ and $\delta$ are roots of $x^2-2(a+1)x+a(a-1)=0$,

$\gamma ,\delta =\frac{2(a+1)\pm \sqrt{4(a+1{)}^2-4a(a-1)}}{2}$

$=\frac{2(a+1)\pm \sqrt{4a^2+8a+4-4a^2+4a}}{2}$

$=\frac{2(a+1)\pm 2\sqrt{3a+1}}{2}$

$=(a+1)\pm \sqrt{3a+1}$

Now, as $\alpha ,\beta \in (\gamma ,\delta )$,

$\alpha ,\beta \in \left(\right(a+1)-\sqrt{3a+1},(a+1)+\sqrt{3a+1})$

It is obvious that $1+a,1-a$ are both $<(a+1)+\sqrt{3a+1}$. (for $a>-\frac{1}{3}$)

So, one condition that needs to be satisfied is $1-a>(a+1)-\sqrt{3a+1}$

$⟹2a<\sqrt{3a+1}$

$⟹4a^2<3a+1$ (squaring both sides)

$⟹4a^2-3a-1<0$

$⟹(4a+1)(a-1)>0$

$⟹a<-\frac{1}{4}$

The other condition that needs to be satisfied is $a+1<(a+1)+\sqrt{3a+1}$

$⟹\sqrt{3a+1}>0$

$⟹a>-\frac{1}{3}$

$\therefore a\in (-\frac{1}{3},-\frac{1}{4})$