Which of the following expressions equals 0 when x < 0 and equals x^2

IMO D

When x<0=> |x|=-x => |x|+x=0
=>[(x+|x|)^2]/4 =0

When x>=0=> |x|=x=> |x|+x=2x
=>[(x+|x|)^2]/4 =(2x)^2/4=x^2

A) If x = -1 < 0, then -1|-1| = -1(1) = -1 ≠ 0.

B) If x = 1 ≥ 0, then 1 + |1| = 2 ≠ 1^2.

C) If x = -1 < 0, then 2(-1)^2 – (-1)|-1| = 2 – (-1) = 2 + 1 = 3 ≠ 0.

E) If x = 1 ≥ 0, then [(1 - |1|)/2]^2 = 0^2 = 0 ≠ 1^2

D) Since we eliminated four incorrect choices, the correct answer must be D.

Answer: D

Note:

Is D really the correct answer? Let’s check.

If x < 0, then (1/4)(x + |x|)^2 = (1/4)(x + (-x))^2 = (1/4)(0)^2 = 0.

If x ≥ 0, then (1/4)(x + |x|)^2 = (1/4)(2x)^2 = (1/4)(4x^2) = x^2.

Hi, I believe there is a slight error:



You mean 3 + 3 = 6

Hi, I believe there is a slight error:



You mean 3 + 3 = 6

Good catch ! Corrected the typo

AryaSwagat your solution is really better and timesaving

 

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Use the definition of absolute values here:
|x| = x if x >= 0
|x| = -x if x < 0

First we check for negative values because we should simply get 0. So we will be able to eliminate many options. 


A. \(x|x|\)
If x < 0, we get -x^2. This is not 0. Eliminate.

B. \(x + |x|\)
If x < 0, we get x - x = 0

C. \(2x^2 - x|x|\)
If x < 0, we get
\(2x^2 + x^2\). This is not 0. Eliminate.

D. \(\frac{1}{4}(x + |x|)^2\)
If x < 0, we get 

\(\frac{1}{4}(x - x)^2 = 0\)

E. \((\frac{x - |x|}{2})^2\)
If x < 0, we get x + x/2. This is not 0. Eliminate.

Now simply check for options (B) and (D).
B. \(x + |x|\)
If x >= 0, we get x + x = 2x. This is not x^2.
Eliminate.

Answer (D)

The definition of absolute values and its usage is discussed here:
https://anaprep.com/algebra-the-why-beh ... questions/