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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
If (x + 2)2 ≤ 2 − y, what is the maximum possible value for y?
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02 Feb 2021, 03:09
02 Feb 2021, 03:09
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If \((x + 2)^2 ≤ 2 − y\), what is the maximum possible value for \(y\)?
A. -4
B. -2
C. 0
D. 2
E. 4
A. -4
B. -2
C. 0
D. 2
E. 4
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1141
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
If (x + 2)2 2 y, what is the maximum possible value for y?
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03 Feb 2021, 09:38
03 Feb 2021, 09:38
2
\((x + 2)^2 ≤ 2 − y\)
Let's simplify this equation and take y to LHS and everything else to RHS
\(y ≤ 2 - (x + 2)^2 \)
Now, y ≤ 2 - square of a number.
And we know that minimum value of square of a number is 0 as square of a number can never be negative.
So, y ≤ 2 - some non-negative number
=> For maximum y, (x + 2)^2 has to be zero
=> Maximum value of y is 2.
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequalities
Let's simplify this equation and take y to LHS and everything else to RHS
\(y ≤ 2 - (x + 2)^2 \)
Now, y ≤ 2 - square of a number.
And we know that minimum value of square of a number is 0 as square of a number can never be negative.
So, y ≤ 2 - some non-negative number
=> For maximum y, (x + 2)^2 has to be zero
=> Maximum value of y is 2.
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequalities
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: If (x + 2)2 ≤ 2 − y, what is the maximum possible value for y?
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26 Sep 2021, 06:46
26 Sep 2021, 06:46
1
\(y ≤ 2 - (x + 2)^2\)
If we consider that inequality takes the equal sign and \(y=0\), then \((x + 2)^2 - 2 = 0\) and \(x^2 + 4x +2 = 0\)
There are two roots and two x-intercepts. The middle between two symmetrical intercepts falls within x for the min/max value of the equation
When \(x=-2\), \(y ≤ 2 - (-2 + 2)^2\) and \(y ≤ 2\)
Out of all answer choices only \(D\) is correct
If we consider that inequality takes the equal sign and \(y=0\), then \((x + 2)^2 - 2 = 0\) and \(x^2 + 4x +2 = 0\)
There are two roots and two x-intercepts. The middle between two symmetrical intercepts falls within x for the min/max value of the equation
When \(x=-2\), \(y ≤ 2 - (-2 + 2)^2\) and \(y ≤ 2\)
Out of all answer choices only \(D\) is correct
GeminiHeat wrote:
If \((x + 2)^2 ≤ 2 − y\), what is the maximum possible value for \(y\)?
A. -4
B. -2
C. 0
D. 2
E. 4
A. -4
B. -2
C. 0
D. 2
E. 4
