If a^2 + 2 b = 7, b^2 + 4 c = -7 and c^2 + 6 a = -14, then what is the value of (a^2 + b^2 + c^2)?
(A) 14
(B) 25
(C) 36
(D) 47
(E) 49
Value of ABC
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- rijul007
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adding all three eqs ..
a^2 + b^2 +c^2 + 6a +2b +4c = -14
a^2 + b^2 +c^2 = -14 - (6a +2b +4c) [even - even = even]
B, D and E are eliminated.
a^2 + 2 b = 7
odd + even = odd
a = odd
b^2 + 4 c = -7
odd + even = odd
b = odd
c^2 + 6 a = -14
even + even = even
c = even
a^2 + b^2 +c^2 = -14 - (6a +2b +4c)
dividing both sides by 2
(a^2 + b^2 +c^2)/2 = -7 - (3a +b +2c) = odd - (odd+odd-even) = odd
(a^2 + b^2 +c^2)/2 = odd
Only A satisfies.
Option A is the correct answer.
a^2 + b^2 +c^2 + 6a +2b +4c = -14
a^2 + b^2 +c^2 = -14 - (6a +2b +4c) [even - even = even]
B, D and E are eliminated.
a^2 + 2 b = 7
odd + even = odd
a = odd
b^2 + 4 c = -7
odd + even = odd
b = odd
c^2 + 6 a = -14
even + even = even
c = even
a^2 + b^2 +c^2 = -14 - (6a +2b +4c)
dividing both sides by 2
(a^2 + b^2 +c^2)/2 = -7 - (3a +b +2c) = odd - (odd+odd-even) = odd
(a^2 + b^2 +c^2)/2 = odd
Only A satisfies.
Option A is the correct answer.
- Tani
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There may be a more sophisticated way to solve this, but the following worked for me.
First I solved for a^2, b^2 and c^2 and added them and got: -14 - 6a - 2b - 4c
That tells me that the answer must be even so it has to be a or c.
Since our answers are integers I assume the variables are integers.
Listing the perfect squares from 1 to 36 I get 1, 4, 9, 16, 25, and 36.
Our task is to find three of then that total to either 14 or 36.
1 + 4 + 9 = 14 so I chose that as my answer.
Checking the answer, (which you don't have to do on test day) we know that the absolute values of a, b, and c are 1, 2, and 3, not necessarily in that order. We also know from our original equations that at least c and a must be negative.
Playing with the possibilities I get a = -3, b = -1 and c = -2
Fooling around with the possibilitie
First I solved for a^2, b^2 and c^2 and added them and got: -14 - 6a - 2b - 4c
That tells me that the answer must be even so it has to be a or c.
Since our answers are integers I assume the variables are integers.
Listing the perfect squares from 1 to 36 I get 1, 4, 9, 16, 25, and 36.
Our task is to find three of then that total to either 14 or 36.
1 + 4 + 9 = 14 so I chose that as my answer.
Checking the answer, (which you don't have to do on test day) we know that the absolute values of a, b, and c are 1, 2, and 3, not necessarily in that order. We also know from our original equations that at least c and a must be negative.
Playing with the possibilities I get a = -3, b = -1 and c = -2
Fooling around with the possibilitie
Tani Wolff
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Add all of the equations together to get:
a^2+6a+b^2+2b+c^2+4c=-14
Complete the square for each variable:
a^2+6a+9 + b^2+2b+1 + c^2+4c+4 = -14+14
(a+3)^2 + (b+1)^2 + (c+2)^2 =0
The sum of three squares can only be zero if all three of them are zero, thus:
a=-3, b=-1. c=-2 and
a^2+b^2+c^2=9+1+4=14
Ans: A
a^2+6a+b^2+2b+c^2+4c=-14
Complete the square for each variable:
a^2+6a+9 + b^2+2b+1 + c^2+4c+4 = -14+14
(a+3)^2 + (b+1)^2 + (c+2)^2 =0
The sum of three squares can only be zero if all three of them are zero, thus:
a=-3, b=-1. c=-2 and
a^2+b^2+c^2=9+1+4=14
Ans: A