Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 0808:00 PM EDT
-10:00 PM EDT
A powerful GMAT course taught live online + 6 months of access to TTP OnDemand GMAT Masterclass included! Mon/Wed June 8, 2026 →August 12, 2026 8:00pm-10:00pm EST
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
Jun 1111:00 AM EDT
-01:00 PM EDT
TTP GMAT OnDemand gives serious students 400+ hours of expert video instruction, the full TTP course, AI support, weekly office hours, and a 715+ score guarantee—all built for elite GMAT score improvement.
01 Jun 2026, 06:57
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
33% (01:24) correct
67%
(02:10)
wrong
based on 43
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are real numbers, then the minimum value of \(x^2+4xy+6y^2-4y+4\) is?
A. -4
B. 0
C. 2
D. 3
E. 4
A. -4
B. 0
C. 2
D. 3
E. 4
01 Jun 2026, 10:51
Kudos
Bookmarks
To find the minimum value of the expression $x^2 + 4xy + 6y^2 - 4y + 4$ for real numbers $x$ and $y$, we can complete the square for both variables.
First, let's group the terms involving $x$ to complete the square for $x$:
$$x^2 + 4xy + 6y^2 - 4y + 4$$
We can split $6y^2$ into $4y^2 + 2y^2$:
$$= (x^2 + 4xy + 4y^2) + 2y^2 - 4y + 4$$
The first group forms a perfect square:
$$= (x + 2y)^2 + 2y^2 - 4y + 4$$
Next, we complete the square for the remaining terms involving $y$:
$$2y^2 - 4y + 4 = 2(y^2 - 2y) + 4$$
$$= 2(y^2 - 2y + 1 - 1) + 4$$
$$= 2(y - 1)^2 - 2 + 4$$
$$= 2(y - 1)^2 + 2$$
Now substitute this back into the expression:
$$= (x + 2y)^2 + 2(y - 1)^2 + 2$$
Since $x$ and $y$ are real numbers, the squared terms $(x + 2y)^2$ and $(y - 1)^2$ are always greater than or equal to zero:
$$(x + 2y)^2 \ge 0$$
$$(y - 1)^2 \ge 0$$
Therefore, the entire expression is minimized when both squared terms equal zero:
$$0 + 2(0) + 2 = 2$$
Correct Option:C. 2
First, let's group the terms involving $x$ to complete the square for $x$:
$$x^2 + 4xy + 6y^2 - 4y + 4$$
We can split $6y^2$ into $4y^2 + 2y^2$:
$$= (x^2 + 4xy + 4y^2) + 2y^2 - 4y + 4$$
The first group forms a perfect square:
$$= (x + 2y)^2 + 2y^2 - 4y + 4$$
Next, we complete the square for the remaining terms involving $y$:
$$2y^2 - 4y + 4 = 2(y^2 - 2y) + 4$$
$$= 2(y^2 - 2y + 1 - 1) + 4$$
$$= 2(y - 1)^2 - 2 + 4$$
$$= 2(y - 1)^2 + 2$$
Now substitute this back into the expression:
$$= (x + 2y)^2 + 2(y - 1)^2 + 2$$
Since $x$ and $y$ are real numbers, the squared terms $(x + 2y)^2$ and $(y - 1)^2$ are always greater than or equal to zero:
$$(x + 2y)^2 \ge 0$$
$$(y - 1)^2 \ge 0$$
Therefore, the entire expression is minimized when both squared terms equal zero:
- $y - 1 = 0 \implies y = 1$
- $x + 2y = 0 \implies x + 2(1) = 0 \implies x = -2$
$$0 + 2(0) + 2 = 2$$
Correct Option:C. 2
architkap
01 Jun 2026, 21:58
Kudos
Bookmarks
We want the minimum value of:
x2 + 4xy + 6y2 − 4y + 4
Step 1: Complete the square for x.
x2 + 4xy + 6y2 − 4y + 4
= (x + 2y)2 + 2y2 − 4y + 4
because
(x + 2y)2 = x2 + 4xy + 4y2.
Step 2: Complete the square for y.
2y2 − 4y + 4
= 2(y2 − 2y) + 4
= 2[(y − 1)2 − 1] + 4
= 2(y − 1)2 + 2
So the expression becomes:
(x + 2y)2 + 2(y − 1)2 + 2
Step 3: Find the minimum.
Both (x + 2y)2 and 2(y − 1)2 are always non-negative.
Therefore, the smallest value occurs when:
x + 2y = 0
and
y = 1
This gives:
x = −2
and
y = 1
Substituting these values makes both square terms 0, so the minimum value is: 2
Answer: 2 ✅
x2 + 4xy + 6y2 − 4y + 4
Step 1: Complete the square for x.
x2 + 4xy + 6y2 − 4y + 4
= (x + 2y)2 + 2y2 − 4y + 4
because
(x + 2y)2 = x2 + 4xy + 4y2.
Step 2: Complete the square for y.
2y2 − 4y + 4
= 2(y2 − 2y) + 4
= 2[(y − 1)2 − 1] + 4
= 2(y − 1)2 + 2
So the expression becomes:
(x + 2y)2 + 2(y − 1)2 + 2
Step 3: Find the minimum.
Both (x + 2y)2 and 2(y − 1)2 are always non-negative.
Therefore, the smallest value occurs when:
x + 2y = 0
and
y = 1
This gives:
x = −2
and
y = 1
Substituting these values makes both square terms 0, so the minimum value is: 2
Answer: 2 ✅
