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Math Revolution GMAT Instructor

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If x+(1/x)=4, x2+(1/x2)=?
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Updated on: 25 Sep 2016, 02:37
Question Stats:

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30% (02:14) wrong

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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, 22:46.

Last edited by

Bunuel on 25 Sep 2016, 02:37, 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)=?
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24 Sep 2016, 23:09
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)=?
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25 Sep 2016, 02:49
MathRevolution wrote:

If x+(1/x)=4, \(x^2+(1/x^2)\)=? A. 10 B. 12 C. 14 D. 16 E. 18

Similar question:

if-x-2-1-x-2-4-what-is-the-value-of-x-4-1-x-175113.html
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Re: If x+(1/x)=4, x2+(1/x2)=?
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25 Sep 2016, 07:59

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)=?
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25 Sep 2016, 08:27
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)=?
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27 Sep 2016, 09:07
==> 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)=?
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17 Sep 2017, 07:58

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)=?
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26 Sep 2017, 23:08

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)=14Hence Answer is option C. Kudos if it helps.

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Re: If x+(1/x)=4, x2+(1/x2)=?
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13 Oct 2018, 02:20

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)?
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24 Jan 2019, 06:30
[

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)?
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24 Jan 2019, 06:45

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)?
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24 Jan 2019, 08:03

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)=?
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27 Jan 2019, 18:26
=>

\(( x + \frac{1}{x} )^2 = x^2 + 2x(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 + (\frac{1}{x})^2 + 2 = 4^2 = 16.\)

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)=?
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27 Jan 2019, 18:26