Quadratic equation, A.P. and G.P. related problem problem

Click For Summary
SUMMARY

The problem involves two quadratic equations, ax² + 2bx + c = 0 and a₁x² + 2b₁x + c₁, which share a common root. It is established that if the ratios a/a₁, b/b₁, and c/c₁ are in arithmetic progression (A.P.), then the coefficients a₁, b₁, and c₁ must be in geometric progression (G.P.). The solution utilizes the properties of A.P. and G.P. to derive the necessary proof through substitution and manipulation of the quadratic formula.

PREREQUISITES
  • Understanding of quadratic equations and their roots
  • Knowledge of arithmetic progression (A.P.) and geometric progression (G.P.)
  • Familiarity with algebraic manipulation and substitution techniques
  • Ability to apply the quadratic formula in problem-solving
NEXT STEPS
  • Study the properties of arithmetic and geometric progressions in depth
  • Learn about the derivation and application of the quadratic formula
  • Explore advanced algebraic techniques for solving polynomial equations
  • Investigate the implications of common roots in polynomial equations
USEFUL FOR

Students studying algebra, mathematicians interested in polynomial relationships, and educators teaching quadratic equations and progressions.

Sumedh
Messages
61
Reaction score
0

Homework Statement


if ax2+2bx+c=0 and a1x2+2b1x+c1 have a common root and
a/a1 ,b/b1 ,c/c1 are in A.P.

show that a1,b1,c1 are in G.P.


Homework Equations





The Attempt at a Solution



I know the mean formula of A.P. i.e. the middle term is the mean of the other two.

any hints of which formula of G.P. to use and how to solve?
 
Physics news on Phys.org
Sumedh said:

Homework Statement


if ax2+2bx+c=0 and a1x2+2b1x+c1 have a common root and
a/a1 ,b/b1 ,c/c1 are in A.P.

show that a1,b1,c1 are in G.P.


Homework Equations





The Attempt at a Solution



I know the mean formula of A.P. i.e. the middle term is the mean of the other two.

any hints of which formula of G.P. to use and how to solve?

You realize that fro an AP, where each term is obtained by adding a fixed amount, you add the 1st and 3rd and divide by 2 to get the 2nd or middle.
In a GP where terms are obtained by multiplying by a fixed amount, you can multiply the 1st an 3rd then take the second root to get the 2nd or middle term. [second root = square root]
 
thank you i got the answer.

let the AP be
(A-D) , (A), (A+D)

then
a=a1(A-D)
b=b1(A)
c=c1(A+D)
one root is common so
putting these values in the formula
(c1a2-c2a1)2 = (a1b2-a2b1)(b1c2-b2c1)

we get the required proof.


thank you very much:smile:
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
16
Views
3K