SOLUTION: if the sum of the roots of quadratic equation, ax^2+bx+c=0, is equal to the sum of cubes of their reciprocals then show that ab^2=3a^2c+c^3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: if the sum of the roots of quadratic equation, ax^2+bx+c=0, is equal to the sum of cubes of their reciprocals then show that ab^2=3a^2c+c^3      Log On


   

Question 850743: if the sum of the roots of quadratic equation, ax^2+bx+c=0, is equal to the sum of cubes of their reciprocals then show that ab^2=3a^2c+c^3
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
I had a couple of typos on the first solution I submitted.
Below I have corrected them.

Edwin

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Question 850743
------------------------
if the sum of the roots of quadratic equation, ax^2+bx+c=0, is equal to the sum of cubes of their reciprocals then show that ab^2=3a^2c+c^3
The sum of the roots of ax%5E2%2Bbx%2Bc%22%22=%22%22%220%22 is -b%2Fa

Use the quadratic formula to find the sum of the cubes of the reciprocals of the roots:

%28%282a%29%2F%28-b+%2B+sqrt%28+b%5E2-4ac+%29%29%29%5E3+%22%22%2B%22%22%28%282a%29%2F%28-b+-+sqrt%28+b%5E2-4ac+%29%29%29%5E3+%22%22=%22%22-b%2Fa

To make things easier, let z%22%22=%22%22sqrt%28+b%5E2-4ac%29

%28%282a%29%2F%28-b+%2B+z%29%29%5E3+%22%22%2B%22%22%28%282a%29%2F%28-b+-+z%29%29%5E3+%22%22=%22%22-b%2Fa

Combine fractions on the left

%288a%5E3%29%2F%28-b+%2B+z%29%5E3+%22%22%2B%22%22%288a%5E3%29%2F%28-b+-+z%29%5E3+%22%22=%22%22-b%2Fa

Divide through by 8a%5E3

1%2F%28-b+%2B+z%29%5E3+%22%22%2B%22%221%2F%28-b+-+z%29%5E3+%22%22=%22%22%28-b%29%2F%288a%5E4%29

%28%28-b+%2B+z%29%5E3%2B%28-b+-+z%29%5E3%29%2F%28%28-b%2Bz%29%5E3%28-b-z%29%5E3%29+%22%22=%22%22%28-b%29%2F%288a%5E4%29

The numerator factors as the sum of two cubes:

 
%28-b%2Bz-b-z%29%28%28b%5E2-2bz%2Bz%5E2%29-%28b%5E2-z%5E2%29%2B%28b%5E2%2B2bz%2Bz%5E2%29%29
%28-2b%29%28b%5E2-2bz%2Bz%5E2-b%5E2%2Bz%5E2%2Bb%5E2%2B2bz%2Bz%5E2%29
%28-2b%29%283z%5E2%2Bb%5E2%29

The denominator simplifies:

%28-b%2Bz%29%5E3%28-b-z%29%5E3
%28%28-b%2Bz%29%28-b-z%29%29%5E3
%28b%5E2-z%5E2%29%5E3

So we have

%28%28-2b%29%283z%5E2%2Bb%5E2%29%29%2F%28b%5E2-z%5E2%29%5E3

Since z%22%22=%22%22sqrt%28+b%5E2-4ac%29
z%5E2%22%22=%22%22+b%5E2-4ac

Substitute for z²

%28%28-2b%29%283%28b%5E2-4ac%29%2Bb%5E2%29%29%2F%28b%5E2-%28b%5E2-4ac%29%29%5E3%22%22=%22%22%28-b%29%2F%288a%5E4%29

%28%28-2b%29%283b%5E2-12ac%2Bb%5E2%29%29%2F%28b%5E2-b%5E2%2B4ac%29%5E3%22%22=%22%22%28-b%29%2F%288a%5E4%29

%28%28-2b%29%284b%5E2-12ac%29%29%2F%284ac%29%5E3%22%22=%22%22%28-b%29%2F%288a%5E4%29 

%28%28-2b%29%284%29%28b%5E2-3ac%29%29%2F%284ac%29%5E3%22%22=%22%22%28-b%29%2F%288a%5E4%29

%28-8b%28b%5E2-3ac%29%29%2F%2864a%5E3c%5E3%29%22%22=%22%22%28-b%29%2F%288a%5E4%29

Simplify the fraction on the left 

%28-b%28b%5E2-3ac%29%29%2F%288a%5E3c%5E3%29%22%22=%22%22%28-b%29%2F%288a%5E4%29

Divide both sides by -b

%28b%5E2-3ac%29%2F%288a%5E3c%5E3%29%22%22=%22%221%2F%288a%5E4%29

Multiply both sides by 8a³

%28b%5E2-3ac%29%2F%28c%5E3%29%22%22=%22%221%2Fa

Cross multiply:

ab%5E2-3a%5E2c%22%22=%22%22c%5E3

ab%5E2%22%22=%22%223a%5E2c%2Bc%5E3

Edwin