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Which of the following is the value of \(\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{2 + \sqrt{3}}\)?
Let's start by understanding what we're being asked to do. We have three fractions that we need to add together:
The challenge here is that each fraction has square roots in the denominator, which makes them difficult to work with directly. Our goal is to find their sum and see which answer choice it matches.
Process Skill: TRANSLATE - Converting the complex fraction sum into manageable pieces
When we have square roots in denominators, we can make them easier to work with by using a technique called "rationalizing." The key idea is simple: if we multiply both the top and bottom of a fraction by the same thing, the fraction's value doesn't change, but we can strategically choose what to multiply by to eliminate the square roots.
For expressions like \((a + b)\), we multiply by \((a - b)\) because \((a + b)(a - b) = a^2 - b^2\).
First fraction: \(\frac{1}{\sqrt{2} + 1}\)
Multiply top and bottom by \((\sqrt{2} - 1)\):
\(= \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)}\)
\(= \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2}\)
\(= \frac{\sqrt{2} - 1}{2 - 1}\)
\(= \sqrt{2} - 1\)
Second fraction: \(\frac{1}{\sqrt{3} + \sqrt{2}}\)
Multiply top and bottom by \((\sqrt{3} - \sqrt{2})\):
\(= \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}\)
\(= \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}\)
\(= \frac{\sqrt{3} - \sqrt{2}}{3 - 2}\)
\(= \sqrt{3} - \sqrt{2}\)
Third fraction: \(\frac{1}{2 + \sqrt{3}}\)
Multiply top and bottom by \((2 - \sqrt{3})\):
\(= \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}\)
\(= \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2}\)
\(= \frac{2 - \sqrt{3}}{4 - 3}\)
\(= 2 - \sqrt{3}\)
Now let's add our three rationalized fractions together:
\((\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3})\)
Something beautiful happens when we look at this sum carefully. Let's group the terms:
\(= \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3}\)
Notice that:
This is what mathematicians call a "telescoping series" - like a telescope collapsing, most terms disappear, leaving us with just:
\(= -1 + 2\)
\(= 1\)
Process Skill: INFER - Recognizing the telescoping pattern that dramatically simplifies the calculation
Let's double-check our work by carefully tracking each term:
Original sum after rationalization:
\((\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3})\)
Rearranging terms by type:
\(= \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} - 1 + 2\)
\(= 0 + 0 + 1\)
\(= 1\)
This confirms that all the middle terms (the square root terms) cancel perfectly, leaving us with just the "end" terms: \(-1 + 2 = 1\).
Looking at our answer choices, this matches choice E exactly.
The value of the given expression is 1, which corresponds to answer choice E.
This elegant result shows the power of recognizing patterns in mathematics - what initially appeared to be a complex calculation involving multiple square roots simplified beautifully through the telescoping property to give us the clean answer of 1.
1. Attempting to find a common denominator instead of rationalizing
Students often see multiple fractions and immediately think they need to find a common denominator to add them. With expressions like \(\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{2 + \sqrt{3}}\), finding a common denominator would lead to an extremely complex expression involving products of terms with multiple square roots. This approach becomes computationally overwhelming and doesn't leverage the elegant telescoping pattern that makes this problem solvable.
2. Not recognizing that rationalization is the key technique needed
Many students get intimidated by square roots in denominators and may try to work with the fractions as-is, or attempt to convert everything to decimals. They miss that rationalizing each fraction individually (multiplying by the conjugate) is the essential first step that transforms this problem from nearly impossible to elegant.
3. Failing to anticipate that the answer might telescope
Students often expect complex problems to have complex intermediate steps and complex final answers. They may not consider that this type of sum might be designed to telescope (cancel most terms), leading them to pursue overly complicated solution paths when a more systematic rationalization approach would reveal the pattern.
1. Making errors in the conjugate multiplication
When rationalizing \(\frac{1}{\sqrt{3} + \sqrt{2}}\), students must multiply by \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\). A common error is using the wrong conjugate (like \(\sqrt{3} + \sqrt{2}\) instead of \(\sqrt{3} - \sqrt{2}\)) or making sign errors in the multiplication. For example, incorrectly calculating \((\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})\) as \(\sqrt{3}^2 + \sqrt{2}^2\) instead of \(\sqrt{3}^2 - \sqrt{2}^2 = 3 - 2 = 1\).
2. Arithmetic mistakes in the difference of squares
Students may correctly identify the conjugates but make calculation errors when applying \((a + b)(a - b) = a^2 - b^2\). For instance, when computing \((\sqrt{2} + 1)(\sqrt{2} - 1)\), they might get \((\sqrt{2})^2 + 1^2 = 2 + 1 = 3\) instead of the correct \((\sqrt{2})^2 - 1^2 = 2 - 1 = 1\).
3. Not carefully tracking positive and negative terms during telescoping
When adding \((\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3})\), students may make sign errors or fail to properly group and cancel terms. They might miss that \(+\sqrt{2}\) and \(-\sqrt{2}\) cancel, or incorrectly handle the \(-1 + 2 = +1\) final calculation, leading to wrong intermediate results.
1. Not trusting the elegant result of 1
After going through complex rationalization steps, many students are surprised that the final answer is the simple integer 1. They may second-guess their work and choose a more "complex-looking" answer like choice A or B, thinking that a problem with so many square roots couldn't possibly simplify to just 1.
2. Mismatching their calculated result with the answer choices
If students made errors in their calculations but arrived at some other clean number (like getting 0, 2, or 3 due to sign errors), they might force a match with the available choices instead of recognizing their error. For example, if they incorrectly got \(\frac{1}{3}\) as their result, they would select choice C without double-checking their telescoping calculation.