Solving for $\sqrt{x} + \frac{1}{\sqrt{x}}$ when $x = 3 - 2\sqrt{2}$

The problem asks us to find the value of the expression $\sqrt{x} + \frac{1}{\sqrt{x}}$ given that $x = 3 - 2\sqrt{2}$. To solve this, we first need to find the value of $\sqrt{x}$.

Finding the Value of $\sqrt{x}$

We are given $x = 3 - 2\sqrt{2}$. We need to find $\sqrt{3 - 2\sqrt{2}}$. Let's try to express $3 - 2\sqrt{2}$ as a perfect square in the form $(a-b)^2$ or $(\sqrt{a}-\sqrt{b})^2$.

Recall the identity: $(\sqrt{a} - \sqrt{b})^2 = (\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 = a + b - 2\sqrt{ab}$.

Comparing $3 - 2\sqrt{2}$ with $a + b - 2\sqrt{ab}$, we can see that we need:

• $a + b = 3$
• $ab = 2$

We need to find two numbers $a$ and $b$ whose sum is 3 and whose product is 2. The numbers 2 and 1 satisfy these conditions ($2+1=3$ and $2 \times 1 = 2$).

So, we can write $3 - 2\sqrt{2}$ as:

$\quad 3 - 2\sqrt{2} = 2 + 1 - 2\sqrt{2 \times 1}$

$\quad 3 - 2\sqrt{2} = (\sqrt{2})^2 + (1)^2 - 2\sqrt{2}\sqrt{1}$

This matches the form $(\sqrt{a} - \sqrt{b})^2$ with $a=2$ and $b=1$.

$\quad 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$

Now we can find $\sqrt{x}$:

$\quad \sqrt{x} = \sqrt{(\sqrt{2} - 1)^2}$

When taking the square root of a square, we get the absolute value:

$\quad \sqrt{(\sqrt{2} - 1)^2} = |\sqrt{2} - 1|$

Since $\sqrt{2} \approx 1.414$, $\sqrt{2} - 1$ is positive ($1.414 - 1 = 0.414 > 0$).

Therefore,

$\quad \sqrt{x} = \sqrt{2} - 1$

Finding the Value of $\frac{1}{\sqrt{x}}$

Now that we have $\sqrt{x} = \sqrt{2} - 1$, we can find $\frac{1}{\sqrt{x}}$:

$\quad \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2} - 1}$

To simplify this expression, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of $\sqrt{2} - 1$, which is $\sqrt{2} + 1$.

$\quad \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$

Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:

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

$\quad \frac{1}{\sqrt{x}} = \frac{\sqrt{2} + 1}{2 - 1}$

$\quad \frac{1}{\sqrt{x}} = \frac{\sqrt{2} + 1}{1}$

$\quad \frac{1}{\sqrt{x}} = \sqrt{2} + 1$

Calculating $\sqrt{x} + \frac{1}{\sqrt{x}}$

Now we have the values for $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$:

• $\sqrt{x} = \sqrt{2} - 1$
• $\frac{1}{\sqrt{x}} = \sqrt{2} + 1$

We can now find their sum:

$\quad \sqrt{x} + \frac{1}{\sqrt{x}} = (\sqrt{2} - 1) + (\sqrt{2} + 1)$

Combine like terms:

$\quad \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{2} + \sqrt{2} - 1 + 1$

$\quad \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{2} + 0$

$\quad \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{2}$

The value of the expression $\sqrt{x} + \frac{1}{\sqrt{x}}$ is $2\sqrt{2}$.

Step-by-Step Calculation Summary

The final answer is $2\sqrt{2}$. This corresponds to option 4.

Revision Table: Square Roots and Expressions

Additional Information: Manipulating Algebraic Expressions

This problem involves simplifying expressions with square roots, which is a common topic in algebra. Recognizing patterns like perfect squares within radicals is key.

• Perfect Squares: $(a+b)^2 = a^2+2ab+b^2$ and $(a-b)^2 = a^2-2ab+b^2$. For radicals, $(\sqrt{a} \pm \sqrt{b})^2 = a \pm 2\sqrt{ab} + b$.
• Conjugates: The conjugate of $p - q\sqrt{r}$ is $p + q\sqrt{r}$. Multiplying a binomial by its conjugate is useful for rationalizing denominators because $(p - q\sqrt{r})(p + q\sqrt{r}) = p^2 - (q\sqrt{r})^2 = p^2 - q^2r$, which is free of radicals if $r$ is a positive integer.
• Simplifying Radicals: Always simplify radicals by factoring out perfect squares, e.g., $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

Understanding these techniques helps in solving a variety of problems involving algebraic expressions with square roots.