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RRB ALP CBT I 17 Feb 2026 Shift 1 Paper

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Question : 50 of 75

Marks: +1, -0

If

4x^{2}+\frac{1}{4x^{2}}=14

, what is the positive value of

x+\frac{1}{4x}

?

1
4
2
8

Solution:

Concept:

Use the algebraic identity

(a+b)^2 = a^2 + b^2 + 2ab

to connect the given expression with

x + \frac{1}{4x}

.

Explanation:

Let

y = 2x + \frac{1}{2x}

.

Squaring both sides:

y^2 = \left(2x + \frac{1}{2x}\right)^2

.

Using the identity:

y^2 = (2x)^2 + \left(\frac{1}{2x}\right)^2 + 2(2x)\left(\frac{1}{2x}\right)

.

This gives

y^2 = 4x^2 + \frac{1}{4x^2} + 2

.

Substitute the given value

4x^2 + \frac{1}{4x^2} = 14

:

y^2 = 14 + 2 = 16

.

Since the positive value is required,

y = \sqrt{16} = 4

.

So

2x + \frac{1}{2x} = 4

.

Dividing both sides by

2

:

x + \frac{1}{4x} = \frac{4}{2} = 2

.

Answer:

The positive value of

x + \frac{1}{4x}

is

2

.

Hence, the correct option is C.

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