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If x+(1/x)=4, x2+(1/x2)=?

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If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post Updated on: 25 Sep 2016, 01:37
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If x+(1/x)=4, \(x^2+(1/x^2)\)=?

A. 10
B. 12
C. 14
D. 16
E. 18

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Originally posted by MathRevolution on 24 Sep 2016, 21:46.
Last edited by Bunuel on 25 Sep 2016, 01:37, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 24 Sep 2016, 22:09
1
MathRevolution wrote:
(ex 12) (equation) If x+(1/x)=4, \(x^2+(1/x^2)\)=?
A. 10 B. 12 C. 14 D. 16 E. 18


x + 1/x = 4

Squaring both sides,

\((x + 1/x )^2 = 4^2\)

or \(x^2+(1/x^2) + 2 = 16\)

or \(x^2+(1/x^2) = 14\)

Hence, C
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 25 Sep 2016, 01:49
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 25 Sep 2016, 06:59
MathRevolution wrote:
If \(x+\frac{1}{x}\) =4, \(x^2+\frac{1}{x^2}\)= ?

A. 10
B. 12
C. 14
D. 16
E. 18


\(x+\frac{1}{x}\) =4

\((x+\frac{1}{x})^2\) = \(4^2\) { Squaring both sides }

Or, \(x^2\) + \(2*x*\frac{1}{x}\)+ \(\frac{1}{x^2}\) = \(16\)

Or, \(x^2\) + \(2\) + \(\frac{1}{x^2}\) = 16

Or, \(x^2+\frac{1}{x^2}\) = \(16 - 2\)

Or, \(x^2+\frac{1}{x^2}\) = \(14\)

So, Answer will be C. 14

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If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 25 Sep 2016, 07:27
Top Contributor
MathRevolution wrote:
If x + (1/x) = 4, \(x^2+(1/x^2)\)=?

A. 10
B. 12
C. 14
D. 16
E. 18


Given: x + (1/x) = 4

NOTE: since there are squared terms in the expression we're trying to evaluate, we should consider squaring both sides.

Square both sides: [x + (1/x)]² = 4²
Rewrite as: [x + (1/x)][x + (1/x)] = 16
Use FOIL method to expand: x² + 1 + 1 + 1/x² = 16
Simplify: x² + 2 + 1/x² = 16
Subtract 2 from both sides: x² + 1/x² = 14

Answer:

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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 27 Sep 2016, 08:07
==> x2+(1/x2)=(x+1/x)2-2x(1/x)=42-2=14 Thus the answer is C.
Answer: C
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 17 Sep 2017, 06:58
MathRevolution wrote:
If x+(1/x)=4, \(x^2+(1/x^2)\)=?

A. 10
B. 12
C. 14
D. 16
E. 18


(x+1/x)^2 = X^2+2+1/x^2;

So x^2+1/x^2 = 16- 2 = 14;

Ans:C
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 26 Sep 2017, 22:08
x + 1/x = 4

Squaring both sides,

(x+1/x)^2 = 4^2

x^2+(1/x^2)+2*x*1/x=16
x^2+(1/x^2)+2=16
x^2+(1/x^2)=14
Hence Answer is option C.

Kudos if it helps.
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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 13 Oct 2018, 01:20
X + 1/x = 4

(x + 1/x)^2 = 4^2

x^2 + 2 (x * 1/x) + 1/x^2 = 16

x^2 + 2 + 1/x^2 = 16

x^2 + 1/x^2 = 16 - 2 = 14

ArjunJag1328 check the above it may clarify your question.

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If x+(1/x)=4, what is the value of x^2+(1/x^2)?  [#permalink]

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New post 24 Jan 2019, 05:30
[Math Revolution GMAT math practice question]

If \(x+(\frac{1}{x})=4\), what is the value of \(x^2+(\frac{1}{x^2})?\)

\(A. 10\)
\(B. 12\)
\(C. 14\)
\(D. 16\)
\(E. 18\)
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Re: If x+(1/x)=4, what is the value of x^2+(1/x^2)?  [#permalink]

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New post 24 Jan 2019, 05:45
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(x+(\frac{1}{x})=4\), what is the value of \(x^2+(\frac{1}{x^2})?\)

\(A. 10\)
\(B. 12\)
\(C. 14\)
\(D. 16\)
\(E. 18\)


Manipulate the if part as

\(x+(\frac{1}{x})^2=16\)

x^2 + 1/x^2 = 16 - 2

= 14

Answer C
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Re: If x+(1/x)=4, what is the value of x^2+(1/x^2)?  [#permalink]

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New post 24 Jan 2019, 07:03
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(x+(\frac{1}{x})=4\), what is the value of \(x^2+(\frac{1}{x^2})?\)

\(A. 10\)
\(B. 12\)
\(C. 14\)
\(D. 16\)
\(E. 18\)


\(x+(\frac{1}{x})=4\)

Squaring both sides we have -

\(x^2 + 2*x*\frac{1}{x} + \frac{1}{x^2} = 16\)

Or, \(x^2 + 2+ \frac{1}{x^2} = 16\)

Or, \(x^2 +\frac{1}{x^2} = 14\), Thus Answer must be (C)

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Re: If x+(1/x)=4, x2+(1/x2)=?  [#permalink]

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New post 27 Jan 2019, 17:26
=>

\(( x + \frac{1}{x} )^2 = x^2 + 2x(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 + (\frac{1}{x})^2 + 2 = 4^2 = 16.\)
Thus, \(x^2 + (\frac{1}{x})^2 = ( x + \frac{1}{x} )^2 - 2 = 16 - 2 = 14.\)

Therefore, the answer is C.
Answer: C
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Re: If x+(1/x)=4, x2+(1/x2)=?   [#permalink] 27 Jan 2019, 17:26
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