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For any two numbers a and b, the operation a#b is given by the expression \(a^2 + 6ab + b^2\). If a#1 = -7, then a could be equal to
(A) -3
(B) -1
(C) -4
(D) 2
(E) 4
_________________
(A) -3
(B) -1
(C) -4
(D) 2
(E) 4
_________________
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IanStewart
Expert
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 10 Sep 2026
Posts: 4,146
Location: United States
GMAT 1: 780 Q51 V47
Expert reply
GMAT 1: 780 Q51 V47
Posts: 4,146
Replacing "b" with "1" in the definition of "a#b", we know we must get "-7" as a result, so:
a^2 + 6(a)(1) + 1^2 = -7
a^2 + 6a + 8 = 0
(a + 4)(a + 2) = 0
and a is either -4 or -2.
_________________
a^2 + 6(a)(1) + 1^2 = -7
a^2 + 6a + 8 = 0
(a + 4)(a + 2) = 0
and a is either -4 or -2.
_________________
now teaching gmat focus
ianstewartgmat.com
ianstewartgmat.com
Posts: 4,849
Putting b = 1, we have a#1 = \(a^2 + 6a + 1 = -7\)
Since \(a^2\) is positive, therefore +6a has to be negative which means a has to be negative.
As per the options either a = -3, -1 or a = -4
Putting a = -3, we get [m]-3^2 + (6 * -3) + 1 = 9 - 18 + 1 = -8 which is not equal to -7
Putting a = -1, we get [m]-1^2 + (6 * -1) + 1 = 1 - 6 + 1 = -4 which is not equal to -7
Putting a = -4, we get [m]-4^2 + (6 * -4) + 1 = 16 - 24 + 1 = -7 which is equal to -7
Therefore a = -4
Option C
Arun Kumar
Since \(a^2\) is positive, therefore +6a has to be negative which means a has to be negative.
As per the options either a = -3, -1 or a = -4
Putting a = -3, we get [m]-3^2 + (6 * -3) + 1 = 9 - 18 + 1 = -8 which is not equal to -7
Putting a = -1, we get [m]-1^2 + (6 * -1) + 1 = 1 - 6 + 1 = -4 which is not equal to -7
Putting a = -4, we get [m]-4^2 + (6 * -4) + 1 = 16 - 24 + 1 = -7 which is equal to -7
Therefore a = -4
Option C
Arun Kumar
Given that a#b = \(a^2 + 6ab + b^2\) and a#1 = -7 and we need to find possible value of of a
To find a#1 we need to compare what is before and after # in a#1 and a#b
=> We need to substitute a with a and b with 1 in a#b to get the value of a#1
=> a#1 = \(a^2 + 6a*1 + 1^2\) = -7 (given)
=> \(a^2 + 6a + 1 + 7\) = 0
=> \(a^2 + 6a + 8\) = 0
=> \(a^2 + 2a + 4a + 8\) = 0
=> a*(a + 2) + 4*(a + 2) = 0
=> (a + 2) * (a + 4) = 0
=> a = -2 or -4
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
_________________
To find a#1 we need to compare what is before and after # in a#1 and a#b
=> We need to substitute a with a and b with 1 in a#b to get the value of a#1
=> a#1 = \(a^2 + 6a*1 + 1^2\) = -7 (given)
=> \(a^2 + 6a + 1 + 7\) = 0
=> \(a^2 + 6a + 8\) = 0
=> \(a^2 + 2a + 4a + 8\) = 0
=> a*(a + 2) + 4*(a + 2) = 0
=> (a + 2) * (a + 4) = 0
=> a = -2 or -4
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
_________________
How to Solve :
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
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