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30 Apr 2024, 16:52
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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64% (01:03) correct
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Which of the following number line graphs represents the solution of the inequality \((x−1)^2≥0\)?
A)
B)
C)
D)
E)
GMAT-Club-Forum-7995eg3l.png [ 42.82 KiB | Viewed 8521 times ]
GMAT-Club-Forum-1szm2rrs.png [ 30.57 KiB | Viewed 8514 times ]
GMAT-Club-Forum-vtko7ts4.png [ 33.54 KiB | Viewed 8527 times ]
GMAT-Club-Forum-9rdz1g6i.png [ 31.95 KiB | Viewed 8510 times ]
GMAT-Club-Forum-dprzdosi.png [ 34.19 KiB | Viewed 8510 times ]
A)
B)
C)
D)
E)
Attachment:
GMAT-Club-Forum-7995eg3l.png [ 42.82 KiB | Viewed 8521 times ]
Attachment:
GMAT-Club-Forum-1szm2rrs.png [ 30.57 KiB | Viewed 8514 times ]
Attachment:
GMAT-Club-Forum-vtko7ts4.png [ 33.54 KiB | Viewed 8527 times ]
Attachment:
GMAT-Club-Forum-9rdz1g6i.png [ 31.95 KiB | Viewed 8510 times ]
Attachment:
GMAT-Club-Forum-dprzdosi.png [ 34.19 KiB | Viewed 8510 times ]
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GMATCoachBen
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GMAT 1: 760 Q50 V44
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GMAT 5: 780 Q51 V48
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30 Apr 2024, 16:56
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GMATCoachBen
What do we know about squares?
A common key property to quickly recognize is that squares are NEVER negative. (Squares are most common, but this is also true for any other even power.)
Therefore, no matter what value x is, the inequality will always be true; \((𝑥−1)^2\) will always be greater than or equal to 0.
Since x could be anything, our answer is A, "all real numbers".
Attachment:
GMAT-Club-Forum-xv6i8v4l.png [ 42.82 KiB | Viewed 8415 times ]
Attachment:
GMAT-Club-Forum-yxg6055c.png [ 30.57 KiB | Viewed 8310 times ]
Attachment:
GMAT-Club-Forum-ua2j4jiy.png [ 33.54 KiB | Viewed 8080 times ]
Attachment:
GMAT-Club-Forum-dz0wb2k6.png [ 31.95 KiB | Viewed 8029 times ]
Attachment:
GMAT-Club-Forum-ed1lher4.png [ 34.19 KiB | Viewed 7977 times ]
01 Aug 2025, 09:24
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GMATCoachBen
Hi GMATCoachBen, please check whether this approach is correct
I am not sure whether this is correct reasoning, but if it is, give a thumb's up !
-> \((x−1)^2≥0\) can also be written as |x-1| ≥ 0
if you solve the above mod.
1] x-1 ≥ 0
= x ≥ 1
2] -(x-1) ≥ 0
= -x+1 ≥ 0
= 1 ≥ x
So x is both greater than equal to 1 and less than equal to 1.
Option A gives this for x
