Question

If the sum of the roots of the quadratic equation \[a{x^2} + bx + c = 0\] is equal to sum of the squares of their reciprocals then \[\dfrac{a}{c},\dfrac{b}{a},\dfrac{c}{b}\] are in
A) AP
B) GP
C) HP
D) None of these

Answer 1

Hint: In this problem, we have to choose the correct progression option from the given options which satisfies the required condition. They give the relation of a quadratic equation with its square of their reciprocals. We are going to find that relation is which progression. Using the given information, at first, we will find the relation of the roots.

Formula used: From the given we will find the nature of the sequence of \[\dfrac{a}{c},\dfrac{b}{a},\dfrac{c}{b}\].
We know that, if three terms \[a,b,c\] are in H.P. then, \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in A.P.
Therefore, the harmonic mean is \[\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\].

Complete step-by-step answer:
It is given that; the sum of the roots of the quadratic equation \[a{x^2} + bx + c = 0\] is equal to the sum of the squares of their reciprocals.
We have to find the nature of \[\dfrac{a}{c},\dfrac{b}{a},\dfrac{c}{b}\].
Let us consider, the roots of \[a{x^2} + bx + c = 0\] are \[\alpha \] and \[\beta \].
Since, the sum of the roots is equal to sum of the squares of their reciprocals, then we have,
\[\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} = \alpha + \beta \]
Simplifying we get,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{{\alpha ^2}{\beta ^2}}} = \alpha + \beta \]
Simplifying again we get,
\[\dfrac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{{\alpha ^2}{\beta ^2}}} = \alpha + \beta \]………………. (A)
Since, \[\alpha \]and \[\beta \] be the roots of \[a{x^2} + bx + c = 0\], then by the relation between roots and coefficient we have,
\[\alpha + \beta = - \dfrac{b}{a}\].................… (1)
\[\alpha \beta = \dfrac{c}{a}\].....................… (2)
Substituting the values of (1) and (2) in (A) we get,
$\Rightarrow$\[\dfrac{{{{\left( {\dfrac{{ - b}}{a}} \right)}^2} - 2\dfrac{c}{a}}}{{\dfrac{{{c^2}}}{{{a^2}}}}} = \dfrac{{ - b}}{a}\]
Simplifying we get,
$\Rightarrow$\[a{b^2} + b{c^2} = 2{a^2}c\]
Simplifying again we get,
$\Rightarrow$\[\dfrac{{2a}}{b} = \dfrac{c}{a} + \dfrac{b}{c}\]
Hence, \[\dfrac{a}{c},\dfrac{b}{a},\dfrac{c}{b}\] are in H.P.

$\therefore $ The correct option is C) H.P.

Note: In this question, we have found the progression of the given relation. An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-one number called the common ratio. A harmonic progression (or harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression.