If $$ \sqrt{x} + \frac{1}{\sqrt{x}} = 7 $$, then the value of $$ x + \frac{1}{x} $$ is equal to:

Question

Accepted Answer

## **Step 1: Let $$ y = \sqrt{x} $$**

So,  
$$
y + \frac{1}{y} = 7
$$

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## **Step 2: Find $$ y^2 + \frac{1}{y^2} $$**

Square both sides:
$$
\left(y + \frac{1}{y}ight)^2 = 7^2 \
y^2 + 2\cdot y \cdot \frac{1}{y} + \frac{1}{y^2} = 49 \
y^2 + 2 + \frac{1}{y^2} = 49 \
y^2 + \frac{1}{y^2} = 49 - 2 = 47
$$

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## **Step 3: Find $$ x + \frac{1}{x} $$**

Recall $$ x = y^2 $$, so:
$$
x + \frac{1}{x} = y^2 + \frac{1}{y^2}
$$

From above,  
$$
x + \frac{1}{x} = 47
$$

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## **Final Answer**

$$
\boxed{47}
$$

Answer

Answer

Answer