Absolute value |x-3|^2 - 4|x-3|=12

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Discussion Overview

The discussion revolves around solving the equation |x-3|^2 - 4|x-3|=12. Participants explore different approaches to solving the equation, including the use of substitution and the implications of absolute values in the context of quadratic equations.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests letting u=|x-3| and derives the quadratic equation u^2 - 4u - 12 = 0, leading to u = 6 and noting that |x-3| = -2 has no solution.
  • Another participant points out that |x-3| = 6 yields two solutions for x, namely x = 9 and x = -3, which was not initially considered.
  • A third participant emphasizes that setting the equation equal to -12 is unnecessary and does not align with the original equation.
  • One participant notes that the absolute value signs can be omitted in the squared term, as squaring eliminates the sign issue, suggesting that (x-3)^2 is equivalent to |x-3|^2.
  • Another participant expresses understanding that the quadratic should be solved first before applying the absolute value conditions.
  • A later reply questions the need for additional work on the negative part of the equation, suggesting that the previous steps were sufficient.

Areas of Agreement / Disagreement

Participants generally agree on the use of substitution and the quadratic approach, but there is disagreement regarding the necessity of considering negative values and the handling of the absolute value in the context of the equation.

Contextual Notes

Some participants highlight the potential confusion around the application of absolute values and the implications of setting the equation equal to negative values, which remains unresolved.

TechnocratX
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Solve |x-3|^2 - 4|x-3|=12

The solution to this equation is -3, 9. But I'm not sure on the working.

There is a hint to let u=|x-3|

So I worked it out the following way.

u^2 -4u = 12

u^2 -4u -12 = 0

(u + 2)(u - 6)=0

u /= -2 and u /= 6

|x-3|=-2 (no solution)
|x-3|= 6, x = 9

Now to work out negative part.

u^2 -4u = -12

u^2 -4u + 12 =0

(no real roots)

So would I use |x-3|=-6, x = -3
 
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TechnocratX said:
Solve |x-3|^2 - 4|x-3|=12

The solution to this equation is -3, 9. But I'm not sure on the working.

There is a hint to let u=|x-3|

So I worked it out the following way.

u^2 -4u = 12

u^2 -4u -12 = 0

(u + 2)(u - 6)=0

u /= -2 and u /= 6

|x-3|=-2 (no solution)
|x-3|= 6, x = 9
You're missing something here.
|x - 3| = 6 means
x - 3 = 6
OR
x - 3 = -6
So you'll get the 2 solutions here.


TechnocratX said:
Now to work out negative part.

u^2 -4u = -12
You don't need this, because the "negative part" was taken care of earlier.
 
In fact, your original question said '= 12'. It makes no sense to set it equal to -12.
 
The u-substitution for |x - 3| should work fine, but another interesting thing to note is that the absolute value signs are not necessary for the first part of the equation. The power of 2 will make anything within the absolute value positive, so |x - 3|^2 is the same things as (x - 3)^2.

Anyways, if you let u = |x - 3| it should be easier.

u^2 - 4u = 12

Then, once you've found u, use the equation u = |x - 3| to solve for x.
 
I understand now. I should just have solved the quadratic, then only applied the absolute value +/- to the u function. Thanks for your help.
 
TechnocratX said:
Now to work out negative part.

u^2 -4u = -12

u^2 -4u + 12 =0

(no real roots)

So would I use |x-3|=-6, x = -3
don't need the extra work when it's already perfect done imho
 

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