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505-555 Level|   Algebra|      
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Bunuel
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 10 Apr 2020, 10:36
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E
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Difficulty:

15% (low)

Question Stats:

82% (01:06) correct 18% (01:41) wrong based on 123 sessions

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If x ≠ 0, then \((x + \frac{1}{x})^2 - (x - \frac{1}{x})^2 =\)

A. 1/x
B. x
C. 1
D. 2
E. 4
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E
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 10 Apr 2020, 10:48
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Bunuel
If x ≠ 0, then \((x + \frac{1}{x})^2 - (x - \frac{1}{x})^2 =\)

A. 1/x
B. x
C. 1
D. 2
E. 4

\((x + \frac{1}{x})^2 - (x - \frac{1}{x})^2 =\)=\((x + \frac{1}{x} + x - \frac{1}{x})*(x + \frac{1}{x} - x + \frac{1}{x}) \)
\((2x)*(\frac{2}{x})\)=4
Ans: E
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 10 Apr 2020, 10:50
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(x+1/x)2−(x−1/x)2
= (x+1/x+x-1/x)(x+1/x-x+1/x)
=2x*2/x
=4 (since x not equal to zero )

correct answer E
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 12 Apr 2020, 02:29
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Solution



Given
In this question, we are given that
    • The number x is not equal to 0

To find
We need to determine
    • The value of the given expression \((x+\frac{1}{x})^2 − (x−\frac{1}{x})^2\)

Approach and Working
Let us assume that \(x + \frac{1}{x}\) as a and \(x – \frac{1}{x}\) as b

So, the given expression
    • \((x+\frac{1}{x})^2 − (x−\frac{1}{x})^2 = a^2 – b^2 = (a + b) * (a – b)\)

Now, let’s substitute the values of a and b
    • \((x + \frac{1}{x} + x – \frac{1}{x}) * (x + \frac{1}{x} – x + \frac{1}{x}) = 2x * \frac{2}{x} = 4\)

Thus, we have simplified the given expression by using the \(a^2 – b^2\) identity

Therefore, option E is the correct answer.

Correct Answer: Option E

To understand how "Simplification" helps you master GMAT Quant, please click on the image below

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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 21 Apr 2020, 09:21
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Hi guys

Any reason why we can't cancel the denominator x by taking it as LCM and multiplying each group with x, leaving us with \((X+1)^2 - (X-1)^2\) and then applying \(x^2+2ab+b^2\) formula on both?

The answer in that case is 2.
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 21 Apr 2020, 09:41
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Hi guys

Any reason why we can't cancel the denominator x by taking it as LCM and multiplying each group with x, leaving us with \((X+1)^2 - (X-1)^2\) and then applying \(x^2+2ab+b^2\) formula on both?

The answer in that case is 2.

you can take the LCM but then you'll have to multiply the equation by x^2.
\(R*x^2 = x^2((\frac{X^2+1}{x})^2 - (\frac{X^2-1}{x})^2)\)
\(R*x^2 = x^2(\frac{(X^2+1)^2}{x^2} - \frac{(X^2-1)^2}{x^2})\)
\(R*x^2 = (X^2+1)^2 - (X^2-1)^2\)
after this you may apply the formula you mentioned or \(a^2-b^2=(a+b)(a-b)\)
it'll give the same answer R =4.
It can be done this way but seems like it'll take a bit more time.
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 21 Apr 2020, 10:23
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EgmatQuantExpert

Solution



Given
In this question, we are given that
    • The number x is not equal to 0

To find
We need to determine
    • The value of the given expression \((x+\frac{1}{x})^2 − (x−\frac{1}{x})^2\)

Approach and Working
Let us assume that \(x + \frac{1}{x}\) as a and \(x – \frac{1}{x}\) as b

So, the given expression
    • \((x+\frac{1}{x})^2 − (x−\frac{1}{x})^2 = a^2 – b^2 = (a + b) * (a – b)\)

Now, let’s substitute the values of a and b
    • \((x + \frac{1}{x} + x – \frac{1}{x}) * (x + \frac{1}{x} – x + \frac{1}{x}) = 2x * \frac{2}{x} = 4\)

Thus, we have simplified the given expression by using the \(a^2 – b^2\) identity

Therefore, option E is the correct answer.

Correct Answer: Option E

To understand how "Simplification" helps you master GMAT Quant, please click on the image below


\((x + \frac{1}{x} + x – \frac{1}{x}) * (x + \frac{1}{x} – x + \frac{1}{x}) = 2x * \frac{2}{x} = 4\)[/list]

How do you get x+1/x in the (a-b) part? Why did the sign change?
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 26 Apr 2020, 04:27
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(x+1/x)2−(x−1/x)2
= 4 *x*1/x
=4

correct answer E
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If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2 = A. 1/x B. x C. 1 D. 2 E. 4 13 Apr 2022, 03:25
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