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For any two numbers a and b, the operation a#b is given by the express

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Bunuel
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For any two numbers a and b, the operation a#b is given by the express [#permalink] New post   27 Jan 2020, 01:57
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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
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C
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Re: For any two numbers a and b, the operation a#b is given by the express [#permalink] New post   27 Jan 2020, 05:43
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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.
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Re: For any two numbers a and b, the operation a#b is given by the express [#permalink] New post   30 Aug 2020, 10:34
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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


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For any two numbers a and b, the operation a#b is given by the express [#permalink] New post   30 Oct 2022, 11:41
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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!

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For any two numbers a and b, the operation a#b is given by the express [#permalink]
30 Oct 2022, 11:41
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