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

[TechnocratX]

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

 

[eumyang]

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.

Now to work out negative part.

u^2 -4u = -12

You don't need this, because the "negative part" was taken care of earlier.

 

[HallsofIvy]

In fact, your original question said '= 12'. It makes no sense to set it equal to -12.

 

[Hertz]

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.

[itex]u^2 - 4u = 12[/itex]

Then, once you've found u, use the equation u = |x - 3| to solve for x.

 

[TechnocratX]

I understand now. I should just have solved the quadratic, then only applied the absolute value +/- to the u function. Thanks for your help.

 

[Latarian]

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