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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 18 Nov 2023, 06:29
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E
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45% (medium)

Question Stats:

68% (01:42) correct 32% (01:35) wrong based on 689 sessions

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What is the sum of the solutions of the equation \((x-1)^2=|x-1|?\)

A. -1
B. 0
C. 1
D. 2
E. 3
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E
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HarshitPanjeta
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 20 Nov 2023, 07:17
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Just drawing a graph will make it very clear that the solutions are 0 and 2. Sum is 2
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 21 Nov 2023, 11:43
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(x-1)^2 = |x-1|, What is x?

(x-1)^2 = x-1 >> x^2-3x+2 = 0 >> x=2, x=1 >> sum of Xs: 3
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 15 Dec 2023, 11:54
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Arghi
(x-1)^2 = |x-1|, What is x?

(x-1)^2 = x-1 >> x^2-3x+2 = 0 >> x=2, x=1 >> sum of Xs: 3
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What about the second case when |x-1| is negative? It becomes 1-x and we would have more x values
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 16 Dec 2023, 08:08
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Arghi
(x-1)^2 = |x-1|, What is x?

(x-1)^2 = x-1 >> x^2-3x+2 = 0 >> x=2, x=1 >> sum of Xs: 3

What about the second case when |x-1| is negative? It becomes 1-x and we would have more x values
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You will also have x^2 - x = 0 which is x=0 or x=1 and x=1 is already one solution of the positive case so you will end up having 0,1,2 to sum up to 3.
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 16 Dec 2023, 08:13
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When |x - 1| is positive

\(x^2-2x+1=x-1\)
\(x^2-3x+2=0\)
\((x-2)(x-1)=0\)

So the 2 roots are x=2 and x=1

When |x - 1| is negative

\(x^2-2x+1=-x+1\)
\(x^2-x=0\)
\(x(x-1)=0\)
The 2 roots here are either x=0 or x=1

Since x=1 is common you just take it once. The sum is 0+1+2= 3
Answer E
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 15 Oct 2024, 19:50
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We need to find What is the sum of all the solutions of \((x-1)^2=|x-1|?\)

We have two cases as we have |x-1|
-Case 1: x-1 ≥ 0 or x ≥ 1

=> | x-1 | = x-1
=> x-1 = \((x-1)^2\)
=> \((x-1)^2\) - (x-1) = 0
=> (x-1)*(x-1 - 1) = 0
=> (x-1)*(x-2) = 0
=> x = 1, 2

But the condition was x ≥ 1 and x = 1 and 2 are both ≥ 1
=> x = 1, 2 are SOLUTIONS		-Case 2: x-1 ≤ 0 or x ≤ 1

=> | x-1 | = -(x-1)
=> -(x - 1) = \((x-1)^2\)
=> \((x-1)^2\) + (x-1) = 0
=> (x-1)*(x-1 + 1) = 0
=> (x-1)*(x)= 0
=> x = 0, 1

But the condition was x ≤ 1 and x = 0 and 1 are both ≤ 1
=> x = 0, 1 are SOLUTIONS

=> Sum of all the solutions for x = 0 + 1 + 2 = 3

So, Answer will be E
Hope it helps!

Watch the following video to MASTER Absolute Values

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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 16 Oct 2024, 19:13
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Im poor with modulus. But when a square value is given equal to a modulus, why do we need to take the case that modulus is negative? isnt it given that x-1 = a square of a vlaue so it should be positive?

Alexabisaad
When |x - 1| is positive

\(x^2-2x+1=x-1\)
\(x^2-3x+2=0\)
\((x-2)(x-1)=0\)

So the 2 roots are x=2 and x=1

When |x - 1| is negative

\(x^2-2x+1=-x+1\)
\(x^2-x=0\)
\(x(x-1)=0\)
The 2 roots here are either x=0 or x=1

Since x=1 is common you just take it once. The sum is 0+1+2= 3
Answer E
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 17 Oct 2024, 00:21
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saurvan
Im poor with modulus. But when a square value is given equal to a modulus, why do we need to take the case that modulus is negative? isnt it given that x-1 = a square of a vlaue so it should be positive?

Alexabisaad
When |x - 1| is positive

\(x^2-2x+1=x-1\)
\(x^2-3x+2=0\)
\((x-2)(x-1)=0\)

So the 2 roots are x=2 and x=1

When |x - 1| is negative

\(x^2-2x+1=-x+1\)
\(x^2-x=0\)
\(x(x-1)=0\)
The 2 roots here are either x=0 or x=1

Since x=1 is common you just take it once. The sum is 0+1+2= 3
Answer E
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|x - 1| cannot be negative, as the absolute value is always more than or equal to 0. The solution you quote incorrectly considers the case when |x - 1| is negative and when |x - 1| is positive. It should consider two cases: when the expression inside the modulus is less than 0 and when it’s more than or equal to 0.

What is the sum of the solutions of the equation \((x-1)^2=|x-1|?\)

A. -1
B. 0
C. 1
D. 2
E. 3

CASE 1:

When x - 1 < 0 (so, when x < 1), |x - 1| = -(x - 1). In this case, we get:

(x - 1)^2 = -(x - 1)

Solving this gives x = 0 or x = 1. Since we are considering the range x < 1, we discard x = 1, so the valid solution for this case is x = 0.

CASE 2:
When x - 1 ≥ 0 (so, when x ≥ 1), |x - 1| = x - 1. In this case, we get:

(x - 1)^2 = x - 1

Solving this gives x = 1 or x = 2. Both solutions are valid in this range.

Therefore, the valid solutions are x = 0, x = 1, and x = 2. The sum of these solutions is 0 + 1 + 2 = 3.

Answer: E


10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.­
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What is the sum of the solutions of the equation (x - 1)^2 = |x - 1|? 17 Nov 2025, 07:36
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