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12 Mar 2020, 01:35
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A
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:
63% (02:18) correct
37%
(02:50)
wrong
based on 317
sessions
History
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Time
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If \(x^2 - 4x + 1 = 0\), what is \(x^2 + \frac{1}{x^2}\)?
A. \(14\)
B. \(16\)
C. \(18\)
D. \(20\)
E. \(22\)
A. \(14\)
B. \(16\)
C. \(18\)
D. \(20\)
E. \(22\)
12 Mar 2020, 02:46
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\(x^2+1 = 4x\)
\(\frac{x^2+1}{x }= 4\)
\(x+\frac{1}{x} = 4\)
squaring both sides
\(x^2 + \frac{1}{x^2} + 2 = 16\)
\(x^2 + \frac{1}{x^2} = 14\)
\(\frac{x^2+1}{x }= 4\)
\(x+\frac{1}{x} = 4\)
squaring both sides
\(x^2 + \frac{1}{x^2} + 2 = 16\)
\(x^2 + \frac{1}{x^2} = 14\)
MathRevolution
General Discussion
16 Mar 2020, 04:05
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=>
\(x^2 - 4x + 1 = 0 \)
\(x – 4 + \frac{1}{x} = 0 \) (dividing both sides by \(x\))
\(x + \frac{1}{x} = 4 \) (adding \(4\) to both sides)
\((\frac{x+1}{x})^2 = 16\) (squaring both sides)
\(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16\) (foiling the left side)
\(x^2 + \frac{1}{x^2} + 2 = 16 \) (simplifying)
\(x^2 + \frac{1}{x^2} = 14 \) (subtracting \(2\) from both sides)
Therefore, A is the answer.
Answer: A
\(x^2 - 4x + 1 = 0 \)
\(x – 4 + \frac{1}{x} = 0 \) (dividing both sides by \(x\))
\(x + \frac{1}{x} = 4 \) (adding \(4\) to both sides)
\((\frac{x+1}{x})^2 = 16\) (squaring both sides)
\(x^2 +2x(\frac{1}{x}) + \frac{1}{x^2} = 16\) (foiling the left side)
\(x^2 + \frac{1}{x^2} + 2 = 16 \) (simplifying)
\(x^2 + \frac{1}{x^2} = 14 \) (subtracting \(2\) from both sides)
Therefore, A is the answer.
Answer: A
