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28 Mar 2016, 22:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:36) correct
42%
(02:32)
wrong
based on 700
sessions
History
Date
Time
Result
Not Attempted Yet
If \(√x -√y+√z=\)0, which of the following is true?
A. \(x^2+y^2+z^2=xy+xz+yz\)
B. \(x^2+y^2+z^2=2xy+2xz+2yz\)
C. \(x^2+y^2-z^2=2xy+2xz-2yz\)
D. \(x^2-y^2-z^2=xy+xz+yz\)
E. \(x^4+y^4+z^4=2xy+2xz+2yz\)
* A solution will be posted in two days.
A. \(x^2+y^2+z^2=xy+xz+yz\)
B. \(x^2+y^2+z^2=2xy+2xz+2yz\)
C. \(x^2+y^2-z^2=2xy+2xz-2yz\)
D. \(x^2-y^2-z^2=xy+xz+yz\)
E. \(x^4+y^4+z^4=2xy+2xz+2yz\)
* A solution will be posted in two days.
25 Dec 2017, 05:49
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MathRevolution
In case one does not remembers the formula, plugging smart numbers will be highly effective.
Let \(x=z=1\) & \(y=4\). this satisfies our stem equation. Test the values in the options and the option where \(LHS=RHS\) will be correct
For Option B \(LHS=1^2+4^2+1^2=18\)
\(RHS=2*1*4+2*1*1+2*4*1=18\).
For other options, LHS will not be equal to RHS
29 Mar 2016, 00:51
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MathRevolution
\(\sqrt{x} + \sqrt{z} = \sqrt{y}\)
\(x + z + 2\sqrt{xz} = y\)
\(x + z - y = -2\sqrt{xz}\)
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
\(x^2 + z^2 + y^2 + 2xz - 2yz - 2xy = 4xz\)
\(x^2 + y^2 + z^2 = 2xy + 2yz + 2xz\)
Answer: B
