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Algebric Question: Absolute Solution Sum

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by MartyMurray » Fri Nov 13, 2015 6:52 am
The problem with doing what you described is that it's not just (x - 3) that we are concerned with.

It's |x - 3|.

So we have two levels we need to solve.

First you need to find the factors that work, the u level factors. Then you need to find all the |x - 3| possibilities that create those u level factors.

Solving for all the possible values of x without substituting u for |x - 3| takes way more work.
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by shahfahad » Fri Nov 13, 2015 6:54 am
Ok thanks :) That was helpful.
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by [email protected] » Fri Nov 13, 2015 9:39 am
Hi shahfahad et al.,

When dealing with these types of complex-looking calculations, sometimes it's easiest to 'rewrite' the given equation.

Notice that |X-3| appears twice in the given equation. If we replace that 'term' with a variable, you should see a common pattern emerge...

Y = |X-3|

With that, the original equation....

|X-3|^2 + |X-3| = 20

Becomes....

Y^2 + Y = 20

From here, we can do Quadratic Algebra and solve for Y...

Y^2 + Y - 20 = 0
(Y+5)(Y-4) = 0

So Y = -5, +4

Going back to the original equation, we now know...
|X-3| = -5 OR |X-3| = +4

Since an absolute value CANNOT have a negative result, we can eliminate the first possibility (-5 is NOT an option) and focus on the second.

|X-3| = +4

X = 7 or -1

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by Matt@VeritasPrep » Fri Nov 13, 2015 2:49 pm
Why not say that

|x-3|² + |x-3| - 20 = 0

(|x-3| + 5)(|x-3| - 4) = 0

|x-3| = -5 (impossible)
or
|x-3| = 4 (x = 7 or x = -1)
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by shahfahad » Sat Nov 14, 2015 3:26 am
Rich.C@EMPOWERgmat.: Thanks Rich for explaining in detail

Matt@VeritasPrep: Nice approach. You made the question look much easier.
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