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 ${x^2} - 2\left( {a + 1} \right)x + a\left( {a - 1} \right) = 0$ such that $\alpha \,,\beta \in \left( {\gamma ,\delta } \right)$, then find the value of ‘a’.

Answer 1

Hint: In this question, first we have to find the value of $\alpha \,,\beta $, then we have to find the value of $\gamma \,,\delta $ and finally we have to apply the given condition i.e.$\alpha \,,\beta \in \left( {\gamma ,\delta } \right)$ to get to the value of a. in this question we will get the value of a in a particular range.

Formula used:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
In the given question, we have
${x^2} - 2x - {a^2} + 1 = 0$
$\Rightarrow {x^2} - 2x - \left( {{a^2} - 1} \right) = 0$
Here, $a = 1\,\,,\,\,b = - 2\,\,,\,\,c = - \left( {{a^2} - 1} \right)$
Now putting these values in formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Also, the two solutions of this equation are $\alpha \,\,and\,\beta $.So,
$\alpha ,\,\beta \, = \,\dfrac{{2 \pm \sqrt {4 - 4 \times 1 \times \left( { - \left( {{a^2} - 1} \right)} \right)} }}{{2 \times 1}}$

On calculation, we get
$\alpha ,\,\beta \, = \,\dfrac{{2 \pm \sqrt {4 + 4\left( {{a^2} - 1} \right)} }}{2}$
On further simplification and taking square root, we get
$ \Rightarrow \dfrac{{2 \pm 2a}}{2}$
$ \Rightarrow 1 \pm a$
Therefore, the values of $\alpha \,,\beta $ are $1 \pm a$.Now, we have
${x^2} - 2\left( {a + 1} \right)x + a\left( {a - 1} \right) = 0$
Here, $a = 1\,\,,\,\,b = - 2\left( {a + 1} \right)\,\,,\,\,c = a\left( {a - 1} \right)$

Now putting these values in formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Also, the two solutions of this equation are $\gamma \,\,and\,\delta $. So,
$\gamma \,,\,\delta \, = \,\dfrac{{2\left( {a + 1} \right) \pm \sqrt {4{{\left( {a + 1} \right)}^2} - 4a\left( {a - 1} \right)} }}{2}$
Now, using formula ${\left( {m + n} \right)^2} = {m^2} + {n^2} + 2mn$
On further calculation, we get
$ \Rightarrow \dfrac{{2\left( {a + 1} \right) \pm \sqrt {4{a^2} + 8a + 4 - 4{a^2} + 4a} }}{2}$

Now, on cancelling $4{a^2}$ and then taking $4$ common and taking it out from the root
$ \Rightarrow \dfrac{{2\left( {a + 1} \right) \pm 2\sqrt {3a + 1} }}{2}$
On dividing, we get
$ \Rightarrow \,\left( {a + 1} \right) \pm \sqrt {3a + 1} $
Therefore, the values of $\gamma \,and\,\delta \,are\,\,\left( {a + 1} \right) \pm \sqrt {3a + 1} $.
Now, as $\alpha \,,\,\beta \, \in \,\left( {\gamma \,,\,\delta } \right)$,
$\alpha ,\beta \in \left( {\left( {a + 1} \right) - \sqrt {3a + 1} ,\left( {a + 1} \right) + \sqrt {3a + 1} } \right)$
It is obvious that $1 + a,1 - a$ are both $ < \,\left( {a + 1} \right) + \sqrt {3a + 1} .\,\left( {for\,a\, > - \dfrac{1}{3}} \right)$.

So, one condition that needs to be satisfied is $1 - a > \left( {a + 1} \right) - \sqrt {3a + 1} $.
$ \Rightarrow 2a < \sqrt {3a + 1} $
Squaring both sides
$ \Rightarrow 4{a^2} < 3a + 1$
$ \Rightarrow 4{a^2} - 3a - 1 < 0$
$ \Rightarrow \left( {4a + 1} \right)\left( {a - 1} \right) > 0$
$ \Rightarrow a < - \dfrac{1}{4}$
The other condition that needs to be satisfied is $a + 1 < \left( {a + 1} \right) + \sqrt {3a + 1} $
$ \Rightarrow \sqrt {3a + 1} > 0$
$ \Rightarrow a > - \dfrac{1}{3}$
$\therefore \,a\, \in \,\left( { - \dfrac{1}{3}, - \dfrac{1}{4}} \right)$

Therefore, the required range for the value of a is $\left( { - \dfrac{1}{3}, - \dfrac{1}{4}} \right)$.

Note: There are basically two roots of a quadratic equation. Here we have found the value of a according to prior restrictions as we have taken in the question and then we have found the final answer considering that restriction and neglecting those which are not satisfying the conditions.