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Algebric Question: Absolute Solution Sum

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by shahfahad » Fri Nov 13, 2015 1:49 am
https://www.gmatprepnow.com/module/gmat- ... video/1018

In this question, if we don't use the "U" replacement method and factor the equation the normal way we do, can't we solve it that way? For example, can't we take x-3 as a common factor from this equation?

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