If \(4x^{2} + \frac{1}{4x^{2}} = 14\), what is the positive value of \(x + \frac{1}{4x}\)?
Algebraic Solution for \(x + \frac{1}{4x}\)
We are given the equation \(4x^{2} + \frac{1}{4x^{2}} = 14\). We need to find the positive value of the expression \(x + \frac{1}{4x}\).
Using Algebraic Identities
Consider the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\). Let's choose \(a = 2x\) and \(b = \frac{1}{2x}\).
- \(a^2 = (2x)^2 = 4x^2\)
- \(b^2 = (\frac{1}{2x})^2 = \frac{1}{4x^2}\)
- \(2ab = 2(2x)(\frac{1}{2x}) = 2\)
Therefore, we can write:
\( \left(2x + \frac{1}{2x}\right)^2 = (2x)^2 + 2(2x)\left(\frac{1}{2x}\right) + \left(\frac{1}{2x}\right)^2 \) \( \left(2x + \frac{1}{2x}\right)^2 = 4x^2 + 2 + \frac{1}{4x^2} \)Rearranging terms, we get:
\( \left(2x + \frac{1}{2x}\right)^2 = \left(4x^2 + \frac{1}{4x^2}\right) + 2 \)Calculating Intermediate Value
Substitute the given value \(4x^2 + \frac{1}{4x^2} = 14\) into the equation:
\( \left(2x + \frac{1}{2x}\right)^2 = 14 + 2 \) \( \left(2x + \frac{1}{2x}\right)^2 = 16 \)To find the value of \(2x + \frac{1}{2x}\), take the square root of both sides:
\( 2x + \frac{1}{2x} = \pm\sqrt{16} \) \( 2x + \frac{1}{2x} = \pm 4 \)Since \(x\) is typically assumed positive in such problems for a positive result, both \(2x\) and \(\frac{1}{2x}\) are positive. Thus, their sum must be positive.
\( 2x + \frac{1}{2x} = 4 \)Finding the Final Value
Now, let's relate the expression we need, \(x + \frac{1}{4x}\), to the value we just found.
Factor out \(\frac{1}{2}\) from the expression \(x + \frac{1}{4x}\):
\( x + \frac{1}{4x} = \frac{1}{2} \left( 2x + \frac{1}{2x} \right) \)Substitute the value \(2x + \frac{1}{2x} = 4\):
\( x + \frac{1}{4x} = \frac{1}{2} (4) \) \( x + \frac{1}{4x} = 2 \)The positive value of \(x + \frac{1}{4x}\) is 2.
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