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If [tex] x + \frac{1}{x} = 2 [/tex], then [tex] \sqrt{x} + \sqrt{\frac{1}{x}} = ? [/tex]

Sagot :

Certainly! Let's solve the given problem step-by-step:

Given the equation [tex]\( x + \frac{1}{x} = 2 \)[/tex].

We want to find the value of [tex]\( \sqrt{x} + \sqrt{\frac{1}{x}} \)[/tex].

Let's define [tex]\( y = \sqrt{x} + \sqrt{\frac{1}{x}} \)[/tex].

Now, we'll square both sides of this equation:

[tex]\[ y^2 = \left(\sqrt{x} + \sqrt{\frac{1}{x}}\right)^2 \][/tex]

By expanding the right side, we get:

[tex]\[ y^2 = (\sqrt{x})^2 + 2 \cdot \sqrt{x} \cdot \sqrt{\frac{1}{x}} + \left(\sqrt{\frac{1}{x}}\right)^2 \][/tex]

Simplifying each term, we have:

[tex]\[ y^2 = x + 2 \cdot \sqrt{x \cdot \frac{1}{x}} + \frac{1}{x} \][/tex]

Given that [tex]\( \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1 \)[/tex], we can further simplify:

[tex]\[ y^2 = x + \frac{1}{x} + 2 \][/tex]

Given in the problem, we know that:

[tex]\[ x + \frac{1}{x} = 2 \][/tex]

Therefore, substituting this back into our equation, we get:

[tex]\[ y^2 = 2 + 2 \][/tex]

Thus:

[tex]\[ y^2 = 4 \][/tex]

To solve for [tex]\( y \)[/tex], we take the square root of both sides:

[tex]\[ y = \sqrt{4} \][/tex]

Hence:

[tex]\[ y = 2 \][/tex]

Therefore, the value of [tex]\( \sqrt{x} + \sqrt{\frac{1}{x}} \)[/tex] is [tex]\( \boxed{2} \)[/tex].