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# What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20?

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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
I didn't think of using substitution on this question. I wanted to do the square of |x-3| but couldn't think of I could do that.
My question: is there a way to multiply two absolute values?
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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
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skylimit wrote:
What is the sum of all possible solutions to $$|x-3|^2 - |x-3| = 20$$?
A) -1
B) 6
C) 7
D) 12
E) 14

Source: GMAT Prep Now - https://www.gmatprepnow.com/module/gmat- ... video/1018

Similar question to practice: what-is-the-sum-of-all-possible-solutions-of-the-equation-x-85988.html
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What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
that's a very tricky one..I started to think of combinations +/+ +/- -/+ -/-, but then noticed that it can be rewritten:
|x-3|^2-20 = |x-3|
it means that |x-3| = 20
now this is more simple!!!
x-3 = 20 -> x = 23
x-3 = -20 -> x=-17
the sum is thus 6.

is my method correct, or just got the right answer by luck?
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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
mvictor wrote:
that's a very tricky one..I started to think of combinations +/+ +/- -/+ -/-, but then noticed that it can be rewritten:
|x-3|^2-20 = |x-3|
it means that |x-3| = 20
now this is more simple!!!
x-3 = 20 -> x = 23
x-3 = -20 -> x=-17
the sum is thus 6.

is my method correct, or just got the right answer by luck?

No, your method is not correct. If you plug in x=23 or x=-17 back to the original equation, you will realize that these are not solutions to the original equation.
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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
GMATinsight wrote:
skylimit wrote:
What is the sum of all possible solutions to $$|x-3|^2 - |x-3| = 20$$?
A) -1
B) 6
C) 7
D) 12
E) 14

Source: GMAT Prep Now - https://www.gmatprepnow.com/module/gmat- ... video/1018

Let, |x-3| = y

i.e. $$|x-3|^2 - |x-3| = 20$$ ---> y^2 - y = 20
i.e. y(y-1) = 20
i.e. Product of two consecutive Numbers = 20
but 5*(5-1)= 20 and also (-4)*(-4-1) = 20
i.e. y = 5 or (-4)
but $$|x-3|$$ can't be Negative i.e. can't be (-4)

Hence, $$|x-3| = 5$$
i.e. x = 8 or -2
Sum of these two solutions = 8+(-2) = 6

(|x-3|^2 - |x-3| = 20)
Let y = |x-3|
So, y^2 - y = 20
y^2 -y - 20 = 0
(y-5)(y+4) = 20

|x-3| = 5,-4
|x-3| = 5 as -4 is -ve and not possible
x-3 = +/- 5
x = 8, -2

Sum of values = 8 -2 = 6
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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
Let |x−3|=Y

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

Y=-4 or 5

Now |x−3|=Y =-4 modulus cannot be negative
|x−3|=Y =5
x-3=5 => x=8
-x+3=5 => x=-2

Sum of the possible solutions = (8-2) = 6

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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
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Re: What is the sum of all possible solutions to |x-3|^2 - |x-3| = 20? [#permalink]
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