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Question

If $(x + \frac{1}{x}) = 7$, then $(x - \frac{1}{x})$ is equal to:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper PDF (31-Jan-2021) (Shift 1)
The correct answer is
$3\sqrt{5}$

The problem asks us to find the value of $(x - \frac{1}{x})$ given that $(x + \frac{1}{x}) = 7$. We can use an algebraic identity relating these two expressions.

Algebraic Identity Application

Consider the squares of the two expressions:

  • $(x + \frac{1}{x})^2 = x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2}$
  • $(x - \frac{1}{x})^2 = x^2 - 2(x)(\frac{1}{x}) + \frac{1}{x^2} = x^2 - 2 + \frac{1}{x^2}$

By comparing these, we can derive the identity:

$(x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4$

Solving for $(x - \frac{1}{x})$

We are given $(x + \frac{1}{x}) = 7$. Substitute this value into the identity:

  1. Square the given value: $(x + \frac{1}{x})^2 = 7^2 = 49$
  2. Use the identity to find $(x - \frac{1}{x})^2$: $(x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4$ $(x - \frac{1}{x})^2 = 49 - 4$ $(x - \frac{1}{x})^2 = 45$
  3. Take the square root of both sides: $(x - \frac{1}{x}) = \pm\sqrt{45}$
  4. Simplify the radical $\sqrt{45}$: $\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$

Therefore, $(x - \frac{1}{x}) = \pm 3\sqrt{5}$. Since the options provided are positive, we select the positive value.

Final Answer

The value of $(x - \frac{1}{x})$ is $3\sqrt{5}$.

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  1. If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:

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