Question

If $\dfrac{1}{{\sqrt \alpha }}$ and $\dfrac{1}{{\sqrt \beta }}$ are the roots of the equation, $a{x^2} + bx + 1 = 0 (a \ne 0,a,b \in R)$, then the equation, $x(x + {b^3}) + ({a^3} - 3abx) = 0$ has roots:-
A. ${\alpha ^{\dfrac{3}{2}}}$ and ${\beta ^{\dfrac{3}{2}}}$
B. $\alpha {\beta ^{\dfrac{1}{2}}}$ and ${\alpha ^{\dfrac{1}{2}}}\beta $
C. $\sqrt {\alpha \beta } $ and $\alpha \beta $
D. ${\alpha ^{\dfrac{{ - 3}}{2}}}$ and ${\beta ^{\dfrac{{ - 3}}{2}}}$

Answer 1

Hint: Here in this question some properties of quadratic equation will get used which are mentioned below:-
*Suppose there is a quadratic equation $a{x^2} + bx + c = 0$ and $\alpha $ and $\beta $ are the two roots then:-
  Sum of roots = \[(\alpha + \beta )\] = $\dfrac{{ - b}}{a}$
  Product of roots = \[(\alpha \beta )\] = $\dfrac{c}{a}$

Complete step-by-step answer:
Quadratic equation given in question is $a{x^2} + bx + 1 = 0 (a \ne 0,a,b \in R)$, so Sum of roots = \[(\alpha + \beta )\] = $\dfrac{{ - b}}{a}$ Equation (1)
Product of roots = \[(\alpha \beta )\] = $\dfrac{1}{a}$ Equation (2)
Also roots of the equation given are $\dfrac{1}{{\sqrt \alpha }}$ and $\dfrac{1}{{\sqrt \beta }}$, so
Sum of roots = $\dfrac{1}{{\sqrt \alpha }} + \dfrac{1}{{\sqrt \beta }}$ Equation (3)
Product of roots = $\dfrac{1}{{\sqrt \alpha }} \times \dfrac{1}{{\sqrt \beta }}$ Equation (4)
Now equating equation 1 with 3 and 2 with 4 we will get,
$ \Rightarrow \dfrac{1}{{\sqrt \alpha }} + \dfrac{1}{{\sqrt \beta }} = \dfrac{{ - b}}{a}$
Taking L.C.M we will get $\dfrac{{\sqrt \beta + \sqrt \alpha }}{{\sqrt \alpha \sqrt \beta }} = \dfrac{{ - b}}{a}$ Equation (5)
$ \Rightarrow \dfrac{1}{{\sqrt \alpha }} \times \dfrac{1}{{\sqrt \beta }} = \dfrac{1}{a}$
We will get $\dfrac{1}{{\sqrt \alpha \sqrt \beta }} = \dfrac{1}{a}$ and after cross multiplying We will get $a = \sqrt \alpha \sqrt \beta $
Now putting this value in equation 5
$ \Rightarrow \dfrac{{\sqrt \beta + \sqrt \alpha }}{a} = \dfrac{{ - b}}{a}$
Cancelling a term from the denominator we will get,
$\therefore \sqrt \beta + \sqrt \alpha = - b$
Now second quadratic equation given is $x(x + {b^3}) + ({a^3} - 3abx) = 0$ which can be further written in general quadratic form as:-
${x^2} + {b^3}x + {a^3} - 3abx = 0$
Now arranging in form of $a{x^2} + bx + c = 0$ we will get,
${x^2} + ({b^3} - 3ab)x + {a^3} = 0$
So the sum and product of roots can be written as:-
Sum of roots = $\dfrac{{ - ({b^3} - 3ab)}}{1}$
Product of roots = $\dfrac{{{a^3}}}{1}$
From above calculation we know that $a = \sqrt \alpha \sqrt \beta $and$\sqrt \beta + \sqrt \alpha = - b$ therefore,
 Product of roots = $\dfrac{{{{(\sqrt \alpha \sqrt \beta )}^3}}}{1} = {\alpha ^{\dfrac{3}{2}}}{\beta ^{\dfrac{3}{2}}}$
Sum of roots = $\dfrac{{{{(\sqrt \beta + \sqrt \alpha )}^3} - 3(\sqrt \beta + \sqrt \alpha )(\sqrt \alpha \sqrt \beta )}}{1}$
$ \Rightarrow \dfrac{{{{(\beta )}^{^{\dfrac{3}{2}}}} + {{(\alpha )}^{\dfrac{3}{2}}}}}{1}$
Therefore roots of the equation$x(x + {b^3}) + ({a^3} - 3abx) = 0$ are ${\alpha ^{\dfrac{3}{2}}}$ and ${\beta ^{\dfrac{3}{2}}}$

So, the correct answer is “Option A”.

Note: Students may likely to make mistakes while calculating the power when square root term is involved so below mentioned one example:-
We know that square root have a power of half so suppose we have one term like this ${(\sqrt \alpha )^5}$ then it will be further simplified like ${({\alpha ^{\dfrac{1}{2}}})^5} = {\alpha ^{\dfrac{5}{2}}}$