
If the sum of the roots of the quadratic equation $ax^{2}+bx+c+0$ is equal to the 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
162k+ views
Hint:To solve the given problem we need to first understand what AP, GP, and HP are.
By applying the formula of the sum of roots and the product of roots we can solve the given problem. We may need to apply algebraic identities while solving the equation.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given quadratic equation is $ax^{2}+bx+c+0$
The sum of the roots are $\alpha +\beta =\dfrac{-b}{a}$
Also,
$\alpha +\beta =\dfrac{1}{\alpha ^{2}}+\dfrac{1}{\beta ^{2}}$
Because according to the question the sum of the roots of the given quadratic equation is equal to the sum of the squares of the reciprocals.
Now,
$\alpha +\beta =\dfrac{\alpha ^{2}+\beta ^{2}}{\left ( \alpha \beta \right )^{2}}$
$ = \dfrac{\left ( \alpha +\beta \right )^{2}-2\left ( \alpha \beta \right )}{\left ( \alpha \beta \right )^{2}}$ $\left [ \alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta \right ]$
The product of the roots are
$\alpha \beta =\dfrac{c}{a}$
Substituting $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ in the above equation, we get
$\dfrac{-b}{a}=\dfrac{\left ( \dfrac{b}{a} \right )^{2}-2\left ( \dfrac{c}{a} \right )}{\left ( \dfrac{c}{a} \right )^{2}}$
Simplifying,
$\dfrac{-b}{a}=\dfrac{b^{2}-2ac}{c^{2}}$
Cross multiplying the above equation, we get
$-bc^{2}=ab^{2}-2a^{2}c$
Now, rearranging the terms of the above equation,
$ab^{2}+bc^{2}=2a^{2}c$
Dividing both sides of the equation by abc, we get
$\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{2a}{b}$
This is true only if $\dfrac{b}{c},\dfrac{a}{b},\dfrac{c}{a}$ are in AP as per the definition of AP.
Hence as per the definition of HP, we can say that their reciprocal $\dfrac{c}{b},\dfrac{b}{a},\dfrac{a}{c}$ are in HP.
Hence, the correct option is (C).
Additional Information:A series of integers in a specified or particular order is referred to as arithmetic progression, or simply as AP.
In a geometric progression, each element following the first is created by multiplying it by a number known as the common ratio, which is represented by the symbol r.
By considering the reciprocals of the arithmetic progression that does not contain zero, a harmonic progression (HP) is described as a sequence of real numbers.
Note: Students must grasp the understanding of the definition of the terms AP, HP, and GP. Only then, they will be able to answer the final result. While solving they must write the equation according to the given condition of the problem.
By applying the formula of the sum of roots and the product of roots we can solve the given problem. We may need to apply algebraic identities while solving the equation.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given quadratic equation is $ax^{2}+bx+c+0$
The sum of the roots are $\alpha +\beta =\dfrac{-b}{a}$
Also,
$\alpha +\beta =\dfrac{1}{\alpha ^{2}}+\dfrac{1}{\beta ^{2}}$
Because according to the question the sum of the roots of the given quadratic equation is equal to the sum of the squares of the reciprocals.
Now,
$\alpha +\beta =\dfrac{\alpha ^{2}+\beta ^{2}}{\left ( \alpha \beta \right )^{2}}$
$ = \dfrac{\left ( \alpha +\beta \right )^{2}-2\left ( \alpha \beta \right )}{\left ( \alpha \beta \right )^{2}}$ $\left [ \alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta \right ]$
The product of the roots are
$\alpha \beta =\dfrac{c}{a}$
Substituting $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ in the above equation, we get
$\dfrac{-b}{a}=\dfrac{\left ( \dfrac{b}{a} \right )^{2}-2\left ( \dfrac{c}{a} \right )}{\left ( \dfrac{c}{a} \right )^{2}}$
Simplifying,
$\dfrac{-b}{a}=\dfrac{b^{2}-2ac}{c^{2}}$
Cross multiplying the above equation, we get
$-bc^{2}=ab^{2}-2a^{2}c$
Now, rearranging the terms of the above equation,
$ab^{2}+bc^{2}=2a^{2}c$
Dividing both sides of the equation by abc, we get
$\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{2a}{b}$
This is true only if $\dfrac{b}{c},\dfrac{a}{b},\dfrac{c}{a}$ are in AP as per the definition of AP.
Hence as per the definition of HP, we can say that their reciprocal $\dfrac{c}{b},\dfrac{b}{a},\dfrac{a}{c}$ are in HP.
Hence, the correct option is (C).
Additional Information:A series of integers in a specified or particular order is referred to as arithmetic progression, or simply as AP.
In a geometric progression, each element following the first is created by multiplying it by a number known as the common ratio, which is represented by the symbol r.
By considering the reciprocals of the arithmetic progression that does not contain zero, a harmonic progression (HP) is described as a sequence of real numbers.
Note: Students must grasp the understanding of the definition of the terms AP, HP, and GP. Only then, they will be able to answer the final result. While solving they must write the equation according to the given condition of the problem.
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