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06 Jan 2022, 04:45
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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:
35%
(medium)
Question Stats:
76% (02:37) correct
24%
(02:57)
wrong
based on 66
sessions
History
Date
Time
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Not Attempted Yet
If \(x^2 − yz = −5, y^2 − xz = 1\) and \(z^2 − xy = 7,\) find the value of \((x − y)^2 + (y − z)^2 + (z − x)^2.\)
(A) 3
(B) 6
(C) 8
(D) 12
(E) 24
(A) 3
(B) 6
(C) 8
(D) 12
(E) 24
06 Jan 2022, 07:14
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Bunuel
Take: \((x − y)^2 + (y − z)^2 + (z − x)^2\)
Expand: \((x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2xz + x^2)\)
Simplify: \(2x^2 - 2xy + 2y^2 - 2yz + 2z^2 - 2xz\)
Factor out of the 2 to get: \(2(x^2 - xy + y^2 - yz + z^2 - xz)\)
Rewrite as follows: \(2[(x^2 - yz) + (y^2 - xz) + (z^2 - xy)]\)
Substitute the given values: \(2[(-5) + (1) + (7)]\)
Evaluate: \(2[3] = 6\)
Answer: B
General Discussion
08 Apr 2023, 05:26
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If you had a quick eye, you would note that each of the terms of \((x-y)^2 + (y-z)^2 + (z-x)^2\) are of the form \((a-b)^2\) which on expansion gives you \( a^2 + b^2 - 2ab\). And thus you would get an \(x^2\), a \(y^2\) and a \(z^2\) and also the negative product of the two variables inside the bracket, viz \(-xy, -zx, \text { and } -yz\), twice. You have all these terms in the following three given equations,
\(x^2 - yz = -5\\
y^2 -xz = 1 \\
z^2 -xy = 7\)
Thus a connection is revealed, and we have to establish the relationship between the above three equations and \((x-y)^2 + (y-z)^2 + (z-x)^2\) .
Now we see that since the variables \(x, y \text{ and } z\) are repeated twice over the three terms, we would have \(2x^2\), \(2y^2\) and \(2z^2\) an also \(-2xz,-2yx \text { and } -2yz\). And all these are presented once in the three given equations, but only once. Thus adding the three equations, we get the expansion of \((x-y)^2 + (y-z)^2 + (z-x)^2\) divided by \(2\), and it adds to \(7 -5 + 1 = 3\). Therefore \((x-y)^2 + (y-z)^2 + (z-x)^2 \text { should be equal to } 2 * 3 = 6. \)
Answer is Choice B.
\(x^2 - yz = -5\\
y^2 -xz = 1 \\
z^2 -xy = 7\)
Thus a connection is revealed, and we have to establish the relationship between the above three equations and \((x-y)^2 + (y-z)^2 + (z-x)^2\) .
Now we see that since the variables \(x, y \text{ and } z\) are repeated twice over the three terms, we would have \(2x^2\), \(2y^2\) and \(2z^2\) an also \(-2xz,-2yx \text { and } -2yz\). And all these are presented once in the three given equations, but only once. Thus adding the three equations, we get the expansion of \((x-y)^2 + (y-z)^2 + (z-x)^2\) divided by \(2\), and it adds to \(7 -5 + 1 = 3\). Therefore \((x-y)^2 + (y-z)^2 + (z-x)^2 \text { should be equal to } 2 * 3 = 6. \)
Answer is Choice B.
