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Let's understand what we're working with here. We have two fractions that we need to add together: \(\frac{\sqrt{2}}{4} + \frac{3}{2\sqrt{2}}\).
The first fraction has a square root in the top (numerator) and a regular number in the bottom (denominator). The second fraction has a regular number in the top but a square root in the bottom. This is like trying to add apples and oranges - we need to get them in the same form first.
Our goal is to combine these into one simplified expression that matches one of the answer choices.
Process Skill: TRANSLATE - Converting the visual form of the fractions into an understanding of what mathematical operations we need to perform
The tricky part here is the second fraction \(\frac{3}{2\sqrt{2}}\) because it has a square root in the denominator. In mathematics, we prefer not to have square roots in the bottom of fractions.
To fix this, we use a technique called "rationalizing the denominator." Think of it like this: if we multiply both the top and bottom of a fraction by the same number, we don't change the fraction's value - we're essentially multiplying by 1.
So let's multiply both top and bottom by \(\sqrt{2}\):
\(\frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{3\sqrt{2}}{2 \times 2} = \frac{3\sqrt{2}}{4}\)
Now our problem becomes: \(\frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{4}\)
This step is much easier now! Looking at our transformed fractions:
\(\frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{4}\)
Both fractions already have the same denominator: 4. This is like having \(\frac{1}{4}\) of a pizza plus \(\frac{3}{4}\) of a pizza - we can simply add the pieces together because they're the same size pieces.
Since both denominators are 4, our common denominator is 4.
Now we can add the fractions by adding the numerators and keeping the common denominator:
\(\frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{4} = \frac{\sqrt{2} + 3\sqrt{2}}{4}\)
In the numerator, we have \(\sqrt{2} + 3\sqrt{2}\). This is like having 1 apple plus 3 apples - we get 4 apples. Similarly, \(\sqrt{2} + 3\sqrt{2} = 4\sqrt{2}\).
So our fraction becomes: \(\frac{4\sqrt{2}}{4} = \sqrt{2}\)
Process Skill: SIMPLIFY - Recognizing that coefficients of like terms can be added, and that \(\frac{4\sqrt{2}}{4}\) simplifies to \(\sqrt{2}\)
Our final answer is \(\sqrt{2}\), which corresponds to answer choice C.
To verify: We started with \(\frac{\sqrt{2}}{4} + \frac{3}{2\sqrt{2}}\), rationalized the second fraction to get \(\frac{\sqrt{2}}{4} + \frac{3\sqrt{2}}{4}\), combined them to get \(\frac{4\sqrt{2}}{4}\), and simplified to \(\sqrt{2}\).
The answer is C.
Many students see the expression \(\frac{\sqrt{2}}{4} + \frac{3}{2\sqrt{2}}\) and attempt to directly add the fractions without addressing the square root in the denominator of the second fraction. They might try to find a common denominator using 4 and \(2\sqrt{2}\) directly, leading to complex and incorrect calculations.
Students often struggle with what to multiply by when rationalizing \(\frac{3}{2\sqrt{2}}\). Some might multiply by \(\sqrt{2}\) only in the numerator or only in the denominator, rather than multiplying both by \(\frac{\sqrt{2}}{\sqrt{2}}\). This fundamental misunderstanding of maintaining equivalent fractions leads to incorrect transformations.
When rationalizing \(\frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\), students frequently make calculation errors. Common mistakes include: getting \(\frac{3\sqrt{2}}{2\sqrt{2}}\) instead of recognizing that \(\sqrt{2} \times \sqrt{2} = 2\), or incorrectly calculating \(2 \times 2 = 4\) in the denominator.
When adding \(\sqrt{2} + 3\sqrt{2}\) in the numerator, students sometimes treat these as unlike terms and leave them uncombined, or incorrectly add them as \(\sqrt{2 + 3} = \sqrt{5}\) instead of recognizing this as \((1 + 3)\sqrt{2} = 4\sqrt{2}\).
Students often fail to recognize that \(\frac{4\sqrt{2}}{4}\) simplifies to \(\sqrt{2}\), either leaving the answer in unreduced form or making errors when canceling the common factor of 4.
No likely faltering points - once students correctly arrive at \(\sqrt{2}\), this directly matches answer choice C without any ambiguity about units, decimal vs. integer forms, or selecting among multiple calculated values.