# Question: Absolute Solution Sum

## Comment on Absolute Solution Sum

### How to solve this question

How to solve this question without the "let something equal" technique?

### It's a lot of work, but here

It's a lot of work, but here it goes....

First recognize that |x-3|² = (x-3)² for all values of x.
So, we have: (x-3)² + |x-3| = 20
Rearrange to get: |x-3| = 20 - (x-3)²
Expand right side: |x-3| = 20 - [x² - 6x + 9]
Simplify right side: |x-3| = -x² + 6x + 11

Now we'll follow the rule from https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid... .

We get two cases:
case 1: x-3 = -x² + 6x + 11
case 2: -(x-3) = -x² + 6x + 11

case 1: x-3 = -x² + 6x + 11
Rearrange to get: x² - 5x - 14 = 0
Factor: (x - 7)(x + 2) = 0
Solve: x = 7 or x = -2

case 2: -(x-3) = -x² + 6x + 11
Simplify left side: -x + 3 = -x² + 6x + 11
Rearrange to get: x² - 7x - 8 = 0
Factor: (x - 8)(x + 1) = 0
Solve: x = 8 or x = -1

So, we have 4 POSSIBLE solutions: x = 7, x = -2, x = 8 and x = -1

At this point, we must TEST each possible solution to check for extraneous roots. When we do so, we find that x = -2 and x = 8 are NOT VALID solutions.

So, the only valid solutions are x = 7 and x = -1
So, the correct answer = 7 + (-1) = 6

### Hi Brent, can you explain how

Hi Brent, can you explain how x=-2 and x = 8 are extraneous roots?

### You bet.

You bet.

The original equation: (x-3)² + |x-3| = 20

Replace x with -2 to get: (-2 - 3)² + |-2 - 3| = 20
Simplify: (-5)² + |-5| = 20
Simplify: 25 + 5 = 20
Doesn't work. So, x = -2 is an extraneous root.

Replace x with 8 to get: (8 - 3)² + |8 - 3| = 20
Simplify: (5)² + |5| = 20
Simplify: 25 + 5 = 20
Doesn't work. So, x = 8 is an extraneous root.

Cheers,
Brent

### very smart approach.

very smart approach. Unfortunately I don't know when to use it :(

### Why did we not take Ix-3I

Why did we not take |x-3| common and then work?

### I mean what would be the

I mean what would be the steps if we are to rewrite the equation as: lx-3l (lx-3l + 1) = 20?

### We COULD rewrite the equation

We COULD rewrite the equation as: |x-3|(|x-3| + 1) = 20, but then we'd be stuck there. We'd be stuck there, because this tells us that something (|x-3|) times something else (|x-3| + 1) equals 20.

Since there are infinitely many ways to have (something)(something else) = 20, we can't do much from this point.

HOWEVER, if we can rewrite the equation as (something)(something else) = 0, (as we do in the video solution), then we know that EITHER (something) = 0 OR (something else) = 0

This is why we need to set quadratic equations equal to zero.

For more on this, watch: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

### Hi

Hi

Why didn't you check for extraneous roots here? I mean why don't we check whether any of 7 or -1 is an extraneous root?

### Good question!

Good question!

In the solution, we looked for extraneous roots indirectly.

When we got to the point where |x - 3| = -5 OR |x - 3| = 4, we recognized that the equation |x - 3| = -5 has no solutions.
And we recognized that the equation |x - 3| = 4 has two good solutions: x = 7 and x = -1.

Since we were able to identify solutions that worked, we didn't need to test for extraneous roots.

Cheers,
Brent

### Hi Brent,

Hi Brent,

So im a bit overwhelmed with the whole concept of inequalities. Just wanted to check with you the 5 main points that I have identified.Please correct me if I am wrong

1. Modulus
For equations like |x-a|=b . we take two conditions of x-a=b and x-a=-b and solve. WE DONT NEED TO PUT BACK THE SOLUTIONS TO CHECK EXTRANEOUS SOLUTIONS.

2. Modulus with variable on both side
For equations like |x-a|=x^2+b . We have to take two conditions and check if the solutions are extraneous by plugging in values and taking only LHS=RHS values

3. Inequalties and absolute value
For equations like -a<|x+a|<a Take the two cases. find the solutions using number line and then take the common solutions. If no common solution then we write the solutions as OR

Equate the equation to 0. Find the solutins and put the solutions in number line. Take 3 different areas and take samplevalues from each of them to check if the inequality is true.

5. We never cancel the common term like y in xy-y^2<0 as we dont know the sign of x unless specified.

### I think I agree with all of

I think I agree with all of these. I might rework point #3 as follows:

3. Inequalities and absolute value (e.g., |x + 3| < k (where k is positive)
So, we get: -k < x + 3 < k
Find the solutions using number line and then take the common solutions.
If no common solution then we write the solutions as OR

The OR here refers to inequalities like |x + 3| > k (where k is positive)
In this case, we get: x + 3 > k OR x + 3 < -k

Cheers,
Brent

### Hi Brent,

Hi Brent,

Am I correct with the below understanding?

That if we square the entity within the modulus then the modulus sign goes away i.e. Can we replace |x-3|^2 with x^2-6x+9?

Warm Regards,
Pritish

### Hi Pritish,

Hi Pritish,

In general, we can say that |something|² = something²

Cheers,
Brent

### Hi Brent,

Hi Brent,

This is how I solved the problem:

|x-3|^2 and |x-3| both are positive
Therefore, |x-3|^2 <20 , i.e, 1,4,9,16
16 is the only option that adds up to 20
So |x-3|=4 ; x=7,-1

Do you think this might be the right approach? Is there a chance to miss something in this process?

Thanks,
S

### That's a clever solution. The

That's a clever solution. The main drawback with that approach is that it assumes that x is an INTEGER.
That said, since all of the answer choices are integers, it's quite likely that x is, indeed, an integer.

Cheers,
Brent

### Ouf! I made the mistake of

Ouf! I made the mistake of taking [x-3]^2 and factoring it out and then solving the equation in which I go x= 7 and -2. I checked -2 in the equation which didn't work so I ended up with just 7, therefore, I chose C. Could you please help explain why "FOILing" out the absolutes value is wrong! Thank you !!!

### Great question!

Great question!
It's true that |x - 3|² = |(x - 3)²| = |x² - 6x + 9|.
However, we can't combine |x² - 6x + 9| and |x - 3| to get: |x² - 5x + 6|

In other words, the following property is NOT necessarily true: |x| + |y| = |x + y|
For example, if x = 2 and y = -2, we get: |2| + |-2| = |2 + (-2)|
Simplify to get: 2 + 2 = |0|, which isn't true.

### Great innovative approach

Great innovative approach Brent. To clarify in this u method, so there's no more the need to test extraneous roots here right? Thanks Brent

### In general, when it comes to

In general, when it comes to absolute value equations, we need only test for extraneous roots when there are VARIABLES on both sides of the equation.
That's said, when solving any equation, it's not a bad idea to confirm that your solution works in the original equation (if you have time of course).

### One more question Brent, we

One more question Brent, we are able to assing u only if both values in absolute are the same as in this case. If it have been |x - 3|² + |x - 2| = 20, does this mean we will need to use usual conventional long method or is there also innovative way of solving like u method too? Thanks Brent

### Yes, to use u-substitution,

Yes, to use u-substitution, the quantities you are replacing with u must all be the same.
Here are a few more questions where this approach has used:
- https://gmatclub.com/forum/what-is-the-sum-of-all-the-real-values-of-x-f...
- https://gmatclub.com/forum/what-is-the-sum-of-all-possible-solutions-of-...
- https://gmatclub.com/forum/what-is-the-sum-of-all-solutions-to-the-equat...

### Hi Brent, I tried solving

Hi Brent, I tried solving this way but got stuck in case 2. What went wrong here? Thanks Brent

|x-3|² + |x-3| = 20
|x-3||x-3|+ |x-3|=20
|x² -6x + 9| + |x-3|=20
|x-²-5x + 6|= 20
x²-5x + 6= 20 or Case2: x²-5x + 6= -20
x²-5x-14=0 x²-5x+26=0
x = 7, -2

### There's no property that says

There's no property that says we can combine terms inside different absolute value expressions.
For example, we can't say that |x²-6x+9|+|x-3|=|x²-5x+6|, as you have done.

Here's a quick example:
Does |-3| + |4| = |(-3) + 4|?
No.
If we evaluate each term, we get: 3 + 4 = 1, which isn't true.

### Great explanation and thanks

Great explanation and thanks Brent. Noted.