SOLUTION: solve for x: 12|1-3x|squared -5|1-3x|-28=0, i think i am thinking too much into this but i don't know need some guidance. thanks if you can get back asap that would be great. than

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Question 246033: solve for x: 12|1-3x|squared -5|1-3x|-28=0, i think i am thinking too much into this but i don't know need some guidance. thanks if you can get back asap that would be great. thank you for all your help.
Found 2 solutions by jsmallt9, Theo:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Assuming that you did not mean to use absolute value symbols, your equation is:
12%281-3x%29%5E2+-+5%281-3x%29+-+28=0

Here's a few ways to solve this:
1) A faster, clever way
This requires that we notice that the expressions in the parentheses are the same. We can replace this expression with a temporary variable:
Let q = 1-3x
Substitute this temporary variable into the equation:
12q%5E2+-+5q+-+28=0
Solve the equation with the temporary variable. Factoring we get:
%283q%2B4%29%284q-7%29+=+0
Now we substitute back in for the temporary variable:
%283%281-3x%29%2B4%29%284%281-3x%29-7%29+=+0
Simplifying we get:
%283-9x%2B4%29%29%284-12x-7%29+=+0
%28-9x%2B7%29%28-12x-3%29=0
By the Zero Product Property this product can be zero only if one of the factors is zero. So
-9x%2B7+=+0 or -12x-3=0
Solving these we get:
-9x+=+-7 or -12x+=+3
x+=+7%2F9 or x+=+%28-3%29%2F12+=+%28-1%29%2F4

2) Even faster and more clever way.
Don't use a temporary variable. Go straight from
12%281-3x%29%5E2+-+5%281-3x%29+-+28=0
to
%283%281-3x%29%2B4%29%284%281-3x%29-7%29+=+0
and finish like above. (This may take some practice to get used to seeing and doing this. But this skill will be useful in many Math classes.)

3) Slowest.
Simplifying 12%281-3x%29%5E2+-+5%281-3x%29+-+28=0
12%281-6x%2B9x%5E2%29+-+5%281-3x%29+-+28=0
12-72x%2B108x%5E2+-+5+%2B+15x+-+28=0
108x%5E2+-+57x+-21+=+0
Factor:
3%2836x%5E2+-+19x+-+7%29+=+0
3%289x+-+7%29%283x+%2B+1%29+=+0
Use the Zero Product Property:
9x+-+7+=+0 or 3x+%2B+1+=+0
9x+=+7 or 3x+=+-1
x+=+7%2F9 or x+=+%28-1%29%2F3

P.S. I hope you will try to learn the faster methods. The "slowest" method is not only slower, but it can be much more difficult because you end up with much larger numbers to work with by multiplying everything out.

P.S.S. Instead of factoring and using the Zero Product Property, you could use the Quadratic Formula instead. I prefer factoring when possible.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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.