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01 May 2019, 00:49
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:28) correct
33%
(02:59)
wrong
based on 319
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x^2 + 4x + 1 = 0\), what is the value of \(x^2 + \frac{1}{x^2}\)?
A. 10
B. 12
C. 14
D. 16
E. 18
A. 10
B. 12
C. 14
D. 16
E. 18
01 May 2019, 01:07
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MathRevolution
i.e. \(x^2 + 4x + 1 = 0\)
Dividing both sides by x we get
i.e. \(x + 4 + \frac{1}{x} = 0\)
i.e. \(x + \frac{1}{x} = -4\)
Squaring both sides we get
\(x^2 + \frac{1}{x^2}+2*x*\frac{1}{x} = 16\)
i.e. \(x^2 + \frac{1}{x^2} = 14\)
Answer: Option C
General Discussion
03 May 2019, 00:16
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=>
\(x^2 + 4x + 1 = 0\)
\(=> x + 4 + \frac{1}{x} = 0\)
\(=> x + \frac{1}{x} = -4\)
So, \((x + \frac{1}{x})^2 = (-4)^2\) and \(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16.\)
Therefore,
\(x^2 +2 + \frac{1}{x^2} = 16\)
and \(x^2 + \frac{1}{x^2} = 14.\)
Therefore, the answer is C.
Answer: C
\(x^2 + 4x + 1 = 0\)
\(=> x + 4 + \frac{1}{x} = 0\)
\(=> x + \frac{1}{x} = -4\)
So, \((x + \frac{1}{x})^2 = (-4)^2\) and \(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16.\)
Therefore,
\(x^2 +2 + \frac{1}{x^2} = 16\)
and \(x^2 + \frac{1}{x^2} = 14.\)
Therefore, the answer is C.
Answer: C
