Question

If $ \alpha ,\beta $ are the roots of $ {{x}^{2}}+px+q=0 $ and also of $ {{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0 $ and if $ \dfrac{\alpha }{\beta },\dfrac{\beta }{\alpha } $ are the roots of $ {{x}^{n}}+1+{{\left( x+1 \right)}^{n}}=0 $ , then n is
(a) an odd integer
(b) an even integer
(c) any integer
(d) none of these

Answer 1

Hint: We should know that the sum of roots are equal to $ -\dfrac{b}{a} $ and the product of roots are equal to $ \dfrac{c}{a} $ . From this, we will get the sum of roots and products of the roots value of equation $ {{x}^{2}}+px+q=0 $ . Then for equation $ {{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0 $ , we can say that $ {{\alpha }^{n}},{{\beta }^{n}} $ are the roots and from this again we can find sum and products of roots. Then in third equation, we will put value of root given i.e. $ \dfrac{\alpha }{\beta } $ and we will get equation as $ {{\left( \dfrac{\alpha }{\beta } \right)}^{n}}+1+{{\left( \dfrac{\alpha }{\beta }+1 \right)}^{n}}=0 $ . On solving this and substituting the values which we got from equation $ {{x}^{2}}+px+q=0 $ and $ {{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0 $ , we will come to know value of n.

Complete step-by-step answer:
Here, we are given that $ \alpha ,\beta $ are the roots of equation $ {{x}^{2}}+px+q=0 $ . We know that sums of roots are equal to $ -\dfrac{b}{a} $ and products of roots are equal to $ \dfrac{c}{a} $ .
So, we can write it from equation $ {{x}^{2}}+px+q=0 $ where a is 1, b is p, c is q. We get as
 $ \alpha +\beta =-p $ ……………………….(1)
 $ \alpha \beta =q $ ………………………………(2)
Similarly, we are given that $ \alpha ,\beta $ are the roots of equation $ {{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0 $ . So, we can say that roots can be written as $ {{\alpha }^{n}},{{\beta }^{n}} $ because x is having power of n.
We can write this equation as
 $ {{\left( {{x}^{n}} \right)}^{2}}+{{p}^{n}}\left( {{x}^{n}} \right)+{{q}^{n}}=0 $
Now, we will assume $ {{x}^{n}}=t $ . So, we get equation after putting the value as
 $ {{t}^{2}}+t{{p}^{n}}+{{q}^{n}}=0 $
So, we can write equation in forms of roots as
 $ {{\left( {{\alpha }^{n}} \right)}^{2}}+{{p}^{n}}\left( {{\alpha }^{n}} \right)+{{q}^{n}}=0 $
 $ {{\left( {{\beta }^{n}} \right)}^{2}}+{{p}^{n}}\left( {{\beta }^{n}} \right)+{{q}^{n}}=0 $
Thus, from above 2 equation, we can say that $ {{\alpha }^{n}},{{\beta }^{n}} $ are roots of quadratic equation $ {{t}^{2}}+t{{p}^{n}}+{{q}^{n}}=0 $ . So, we get as
Sum of roots $ {{\alpha }^{n}}+{{\beta }^{n}}=-{{p}^{n}} $ …………………………..(3)
Product of roots $ {{\alpha }^{n}}{{\beta }^{n}}={{q}^{n}} $ ……………………………(4)
Now, we are given that $ \dfrac{\alpha }{\beta },\dfrac{\beta }{\alpha } $ are the roots of $ {{x}^{n}}+1+{{\left( x+1 \right)}^{n}}=0 $ . So, putting the root $ \dfrac{\alpha }{\beta } $ in equation, we get equation as
 $ {{\left( \dfrac{\alpha }{\beta } \right)}^{n}}+1+{{\left( \dfrac{\alpha }{\beta }+1 \right)}^{n}}=0 $
On further solving, we get as
 $ \dfrac{{{\alpha }^{n}}}{{{\beta }^{n}}}+1+\dfrac{{{\left( \alpha +\beta \right)}^{n}}}{{{\beta }^{n}}}=0 $
On taking LCM, we can write it as
 $ \dfrac{{{\alpha }^{n}}+{{\beta }^{n}}+{{\left( \alpha +\beta \right)}^{n}}}{{{\beta }^{n}}}=0 $
On putting value of equation (1) and (3) in above equation, we get as
 $ -{{p}^{n}}+{{\left( -p \right)}^{n}}=0 $
So, further we get as
 $ {{\left( -p \right)}^{n}}={{p}^{n}} $
We can write $ -p=\left( -1 \right)p $ . So, we get as
 $ {{\left( -1 \right)}^{n}}{{p}^{n}}={{p}^{n}} $
On cancelling the same terms, we get as
 $ {{\left( -1 \right)}^{n}}=1 $
So, from the above equation we can know that if n is an even number then, we will get positive number 1 on the right hand side.
Thus, n is an even integer.
Hence, option (b) is the correct answer.

Note: Be careful in this type of problem. We are asked to find values of n and not values of roots. Sometimes students first find values of roots and the problem becomes very difficult to solve. Also, students should know the sum of roots are equal to $ -\dfrac{b}{a} $ and product of roots are equal to $ \dfrac{c}{a} $ otherwise the problem will not be solved properly.