Question 246033
12*|1-3x|^2 - 5*|1-3x| - 28 = 0


let y = |1-3x|


your equation becomes:


12*y^2 - 5*y - 28 = 0


use the quadratic formula to solve.


you get:


y = 1.75 or y = -1.33333333


Assume y = 1.75


This means that |1-3x| = 1.75


If the expression within the absolute value sign is positive, this means that:


1-3x = 1.75


Solve for x to get x = -.25


Substitute in the original equation of |1-3x| = 1.75 to confirm (it does).


If the expression within the absolute value sign is negative, this means that:


-(1-3x) = 1.75


multiply both sides of this equation by -1 to get:


1-3x = -1.75


Solve for x to get x = .9166666667


Substitute in the original equation of |1-3x| = 1.75 to confirm (it does).


Assume y = -1.33333333


This makes |1-3x| = -1.333333333


Since the absolute value of a number can never be negative, this solution is rejected as being invalid.


You have 2 possible answers.


They are:


x = -.25


x = .9166666667


Since |1-3x| is equal to 1.75 for both of these numbers, then use |1-3x| = 1.75 in your original equation to confirm.


12*(1.75)^2 - 5*(1.75) - 28 = 0 becomes 0 = 0 confirming that the 2 values of x you obtained earlier are correct.