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a,b and c are distinct numbers such that abc≠0
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Updated on: 12 Jul 2020, 11:05
Updated on: 12 Jul 2020, 11:05
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Question Stats:
47% (01:16) correct
52% (00:44) wrong based on 38 sessions
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a,b and c are distinct numbers such that abc≠0
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
Quantity A |
Quantity B |
a^2 + b^2 + c^2 - ab -bc - ca |
0 |
My channel URL: https://www.youtube.com/channel/UCvdY0k ... EOsT5PMFRQ
Re: abc≠0
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12 Jul 2020, 10:53
12 Jul 2020, 10:53
1
I might be wrong, but it seems like the answer imbedded in the question is wrong based on the specified assumptions
a^2 + b^2 + c^2 - ab -bc - ca
First we can combine a^2 and b^2 together using the square of difference (a-b)^2 = a^2 -2ab +b^2
Thus a^2 + b^2 - ab part of the equation can be expressed as (a-b)^2 + ab (the plus ab term is needed to make -2ab equal to -ab)
Then we are left with: (a-b)^2 + ab -bc - ca + c^2
This can be rewritten as: (a-b)^2 + b(a -c) + c(c-a)
Rewrite c(c-a) as -c(a-c) ; yielding (a-b)^2 + b(a -c) - c(a-c)
Since the second and third term of the equation contain a common element (a-c) we can further rewrite as:
(a-b)^2 + (b-c)(a -c)
Now, the (a-b)^2 is always positive regardless of values of a or b (because squared)
(b-c) and (a-c) terms can also be positive when a and b are larger than c
However, nowhere in the problem is it indicated that a, b, and c are distinct numbers, or that they cannot equal to each other.
Thus one of the solutions to the equation is the scenario where a=b=c
Under these conditions the entire polynomial will equal to zero
Thus without additional specifications, the answer should be D (not enough info provided)
a^2 + b^2 + c^2 - ab -bc - ca
First we can combine a^2 and b^2 together using the square of difference (a-b)^2 = a^2 -2ab +b^2
Thus a^2 + b^2 - ab part of the equation can be expressed as (a-b)^2 + ab (the plus ab term is needed to make -2ab equal to -ab)
Then we are left with: (a-b)^2 + ab -bc - ca + c^2
This can be rewritten as: (a-b)^2 + b(a -c) + c(c-a)
Rewrite c(c-a) as -c(a-c) ; yielding (a-b)^2 + b(a -c) - c(a-c)
Since the second and third term of the equation contain a common element (a-c) we can further rewrite as:
(a-b)^2 + (b-c)(a -c)
Now, the (a-b)^2 is always positive regardless of values of a or b (because squared)
(b-c) and (a-c) terms can also be positive when a and b are larger than c
However, nowhere in the problem is it indicated that a, b, and c are distinct numbers, or that they cannot equal to each other.
Thus one of the solutions to the equation is the scenario where a=b=c
Under these conditions the entire polynomial will equal to zero
Thus without additional specifications, the answer should be D (not enough info provided)
Re: a,b and c are distinct numbers such that abc≠0
[#permalink]
12 Jul 2020, 16:32
12 Jul 2020, 16:32
1
The answer is D.
a^2 + b^2 + c^2 - ab -bc - ca = (a+b+c)^2- 3(ab+bc+ca)
Put, a= 0.1, b= 0.2, c=0.3, then this becomes negative. So, Quantity B is greater
Put, a= 1, b= 5, c= 6, then this becomes positive. So, Quantity A is greater
Finally, answer is D
[Do not advertise yourself with wrong questions and answers please]
a^2 + b^2 + c^2 - ab -bc - ca = (a+b+c)^2- 3(ab+bc+ca)
Put, a= 0.1, b= 0.2, c=0.3, then this becomes negative. So, Quantity B is greater
Put, a= 1, b= 5, c= 6, then this becomes positive. So, Quantity A is greater
Finally, answer is D
[Do not advertise yourself with wrong questions and answers please]
