Quadratic Equation - Roots for non-zero denominator example ( Enter your problem )
  1. Example-1 : `(5X-18)/(X+2) = (2X-6)/(X-1)`
  2. Example-2 : `(x)/(x+1) + (x+1)/(x) = 5/2`
  3. Example-3 : `4((4x+1)/(4x-1))^(2) + (4x+1)/(4x-1) = 3`
  4. Example-4 : `(4x+1)/(4x-1) + (4x-1)/(4x+1) = 3`

2. Example-2 : `(x)/(x+1) + (x+1)/(x) = 5/2`
(Previous example)
4. Example-4 : `(4x+1)/(4x-1) + (4x-1)/(4x+1) = 3`
(Next example)

3. Example-3 : `4((4x+1)/(4x-1))^(2) + (4x+1)/(4x-1) = 3`





3. Find roots of the equation `12((2X+1)/(X-1))^2 + -5((2X+1)/(X-1)) + -2 = 0`

` 12((2X+1)/(X-1))^2 - 5((2X+1)/(X-1)) - 2 = 0`

` "Let " (2X+1)/(X-1) = m`

` => (12m^2-5m-2) = 0`

` => 12m^2-5m-2 = 0`

` => (12m^2-5m-2) = 0`

` => (12m^2+3m-8m-2) = 0`

` => 3m(4m+1)+(-2)(4m+1) = 0`

` => (3m-2)(4m+1) = 0`

` => (3m-2) = 0" or "(4m+1) = 0`

` => 3m = 2" or "4m = -1`

` => m = 2/3" or "m = -1/4`

` "Now, " (2X+1)/(X-1) = 2/3`

` => 3(2X+1) = 2(X-1)`

` => 3(2X+1) - 2(X-1) = 0`

` => (3(2X+1)-2(X-1)) = 0`

` => (4X+5) = 0`

` => 4X = -5`

` => X = -5/4`

` "Now, " (2X+1)/(X-1) = -1/4`

` => 4(2X+1) = -1(X-1)`

` => 4(2X+1) + 1(X-1) = 0`

` => (4(2X+1)+(X-1)) = 0`

` => (9X+3) = 0`

` => 9X = -3`

` => X = -3/9`

` => X = -1/3`




This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



2. Example-2 : `(x)/(x+1) + (x+1)/(x) = 5/2`
(Previous example)
4. Example-4 : `(4x+1)/(4x-1) + (4x-1)/(4x+1) = 3`
(Next example)





Share this solution or page with your friends.


 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.