Question

If in the equation \[a{x^2} + bx + c = 0\], the sum of roots is equal to sum of square of their reciprocals, then \[\frac{c}{a},\frac{a}{b},\frac{b}{c}\] are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint
The product of its roots is equal to \[c/a\] and the sum of its roots is \[ - b/a\] for the quadratic equation \[a{x^2} + bx + c{\rm{ }} = {\rm{ }}0\]. The product of two binomials can be used to represent a quadratic equation. The roots of the given quadratic equation in this case are a and b. The "solutions" or numerical values that are equal to the given equation's variable are known as the roots of the equation.
The x-intercepts of a function are its roots. The y-coordinate of a point on the x-axis is always zero. The square root of a number is that factor, and the original value is obtained by multiplying the square root by itself.
Formula use:
The arithmetic progression is \[a,b,c\]
\[(b - a) = (c - b)\]
if he roots of the equation \[a{x^2} + bx + c{\rm{ }} = {\rm{ }}0\] be p and q.
The sum of roots is \[p + q = - \frac{b}{a}\]and
The product of roots is \[pq = \frac{c}{a}\]
Complete step-by-step solution
Assume that the roots of the equation \[a{x^2} + bx + c{\rm{ }} = {\rm{ }}0\]be p and q.
The sum of roots is \[p + q = - \frac{b}{a}\]and
The product of roots is \[pq = \frac{c}{a}\]
Given that the reciprocal sum of the squares of the roots' reciprocals equals the total of the roots,
\[p + q = \frac{1}{{{p^2}}} + \frac{1}{{{q^2}}}\]
The equation can be simplified as
\[p + q = \frac{{{q^2} + {p^2}}}{{{p^2}{q^2}}}\]
\[p + q = \frac{{{{(q + p)}^2} - 2pq}}{{{{(pq)}^2}}}\]
\[ - \frac{b}{a} = \frac{{{{( - \frac{b}{a})}^2} - 2\frac{c}{a}}}{{{{(\frac{c}{a})}^2}}}\]
Then the equation is simplified and calculated as
\[ = > b{c^2} = - a{b^2} + 2{a^2}c\]
On both sides, divide the equation by \[abc\]
\[ = > \frac{{b{c^2}}}{{abc}} = \frac{{ - a{b^2} + 2{a^2}c}}{{abc}}\]
\[ = > \frac{c}{a} + \frac{b}{c} = 2\frac{a}{b}\]
So, these are determined to be in the A.P series.
Therefore, the correct option is A.

Note
Numbers' square sums are referred to as the sum of squares. In essence, it is just adding squared numbers. To calculate RSS, first subtract the actual observed values from the estimated values to get the model's level of error or residue. To get RSS, square all error numbers and put them together. The accuracy of the regression prediction improves with decreasing model error.