Evaluating Expressions with Square Roots: A Step-by-Step Solution

The problem asks us to find the value of a complex expression involving several terms with square roots in the denominator and alternating signs. The expression is given by:

\(\frac{1}{{4 - \sqrt {15} }} - \frac{1}{{\sqrt {15} - \sqrt {14} }} + \frac{1}{{\sqrt {14} - \sqrt {13} }} - \frac{1}{{\sqrt {13} - \sqrt {12} }} + \frac{1}{{\sqrt {12} - \sqrt {11} }} - \frac{1}{{\sqrt {11} - \sqrt {10} }} + \frac{1}{{\sqrt {10} - 3}} - \frac{1}{{3 - \sqrt 8 }}\)

To simplify this expression, we will rationalize the denominator of each term individually. Rationalizing the denominator involves multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of \(a - \sqrt{b}\) is \(a + \sqrt{b}\), and the conjugate of \(\sqrt{a} - \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\). This uses the difference of squares formula: \((x-y)(x+y) = x^2 - y^2\).

Step-by-Step Rationalization

Let's rationalize each term:

1. First Term: \(\frac{1}{{4 - \sqrt {15} }}\)

   Multiply numerator and denominator by the conjugate, \(4 + \sqrt{15}\):

   \(\frac{1}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }} = \frac{{4 + \sqrt {15} }}{{4^2 - (\sqrt {15} )^2}} = \frac{{4 + \sqrt {15} }}{{16 - 15}} = \frac{{4 + \sqrt {15} }}{1} = 4 + \sqrt {15} \)

2. Second Term: \(\frac{1}{{\sqrt {15} - \sqrt {14} }}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{15} + \sqrt{14}\):

   \(\frac{1}{{\sqrt {15} - \sqrt {14} }} \times \frac{{\sqrt {15} + \sqrt {14} }}{{\sqrt {15} + \sqrt {14} }} = \frac{{\sqrt {15} + \sqrt {14} }}{{(\sqrt {15} )^2 - (\sqrt {14} )^2}} = \frac{{\sqrt {15} + \sqrt {14} }}{{15 - 14}} = \frac{{\sqrt {15} + \sqrt {14} }}{1} = \sqrt {15} + \sqrt {14} \)

   The expression has \(-\frac{1}{{\sqrt {15} - \sqrt {14} }}\), so this term is \( -(\sqrt {15} + \sqrt {14} ) = -\sqrt {15} - \sqrt {14} \).

3. Third Term: \(\frac{1}{{\sqrt {14} - \sqrt {13} }}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{14} + \sqrt{13}\):

   \(\frac{1}{{\sqrt {14} - \sqrt {13} }} \times \frac{{\sqrt {14} + \sqrt {13} }}{{\sqrt {14} + \sqrt {13} }} = \frac{{\sqrt {14} + \sqrt {13} }}{{(\sqrt {14} )^2 - (\sqrt {13} )^2}} = \frac{{\sqrt {14} + \sqrt {13} }}{{14 - 13}} = \frac{{\sqrt {14} + \sqrt {13} }}{1} = \sqrt {14} + \sqrt {13} \)

4. Fourth Term: \(\frac{1}{{\sqrt {13} - \sqrt {12} }}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{13} + \sqrt{12}\):

   \(\frac{1}{{\sqrt {13} - \sqrt {12} }} \times \frac{{\sqrt {13} + \sqrt {12} }}{{\sqrt {13} + \sqrt {12} }} = \frac{{\sqrt {13} + \sqrt {12} }}{{(\sqrt {13} )^2 - (\sqrt {12} )^2}} = \frac{{\sqrt {13} + \sqrt {12} }}{{13 - 12}} = \frac{{\sqrt {13} + \sqrt {12} }}{1} = \sqrt {13} + \sqrt {12} \)

   The expression has \(-\frac{1}{{\sqrt {13} - \sqrt {12} }}\), so this term is \( -(\sqrt {13} + \sqrt {12} ) = -\sqrt {13} - \sqrt {12} \).

5. Fifth Term: \(\frac{1}{{\sqrt {12} - \sqrt {11} }}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{12} + \sqrt{11}\):

   \(\frac{1}{{\sqrt {12} - \sqrt {11} }} \times \frac{{\sqrt {12} + \sqrt {11} }}{{\sqrt {12} + \sqrt {11} }} = \frac{{\sqrt {12} + \sqrt {11} }}{{(\sqrt {12} )^2 - (\sqrt {11} )^2}} = \frac{{\sqrt {12} + \sqrt {11} }}{{12 - 11}} = \frac{{\sqrt {12} + \sqrt {11} }}{1} = \sqrt {12} + \sqrt {11} \)

6. Sixth Term: \(\frac{1}{{\sqrt {11} - \sqrt {10} }}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{11} + \sqrt{10}\):

   \(\frac{1}{{\sqrt {11} - \sqrt {10} }} \times \frac{{\sqrt {11} + \sqrt {10} }}{{\sqrt {11} + \sqrt {10} }} = \frac{{\sqrt {11} + \sqrt {10} }}{{(\sqrt {11} )^2 - (\sqrt {10} )^2}} = \frac{{\sqrt {11} + \sqrt {10} }}{{11 - 10}} = \frac{{\sqrt {11} + \sqrt {10} }}{1} = \sqrt {11} + \sqrt {10} \)

   The expression has \(-\frac{1}{{\sqrt {11} - \sqrt {10} }}\), so this term is \( -(\sqrt {11} + \sqrt {10} ) = -\sqrt {11} - \sqrt {10} \).

7. Seventh Term: \(\frac{1}{{\sqrt {10} - 3}}\)

   Multiply numerator and denominator by the conjugate, \(\sqrt{10} + 3\):

   \(\frac{1}{{\sqrt {10} - 3}} \times \frac{{\sqrt {10} + 3}}{{\sqrt {10} + 3}} = \frac{{\sqrt {10} + 3}}{{(\sqrt {10} )^2 - 3^2}} = \frac{{\sqrt {10} + 3}}{{10 - 9}} = \frac{{\sqrt {10} + 3}}{1} = \sqrt {10} + 3\)

8. Eighth Term: \(\frac{1}{{3 - \sqrt 8 }}\)

   Multiply numerator and denominator by the conjugate, \(3 + \sqrt{8}\):

   \(\frac{1}{{3 - \sqrt 8 }} \times \frac{{3 + \sqrt 8 }}{{3 + \sqrt 8 }} = \frac{{3 + \sqrt 8 }}{{3^2 - (\sqrt 8 )^2}} = \frac{{3 + \sqrt 8 }}{{9 - 8}} = \frac{{3 + \sqrt 8 }}{1} = 3 + \sqrt 8 \)

   The expression has \(-\frac{1}{{3 - \sqrt 8 }}\), so this term is \( -(3 + \sqrt 8 ) = -3 - \sqrt 8 \). Also, \(\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}\). So this term is \( -3 - 2\sqrt{2} \).

Summing the Rationalized Terms

Now, let's combine all the simplified terms according to the original expression:

\((4 + \sqrt {15} ) + (-\sqrt {15} - \sqrt {14} ) + (\sqrt {14} + \sqrt {13} ) + (-\sqrt {13} - \sqrt {12} ) + (\sqrt {12} + \sqrt {11} ) + (-\sqrt {11} - \sqrt {10} ) + (\sqrt {10} + 3) + (-3 - 2\sqrt 2 )\)

Let's group the terms and see the cancellations:

\(4 + \sqrt {15} - \sqrt {15} - \sqrt {14} + \sqrt {14} + \sqrt {13} - \sqrt {13} - \sqrt {12} + \sqrt {12} + \sqrt {11} - \sqrt {11} - \sqrt {10} + \sqrt {10} + 3 - 3 - 2\sqrt 2 \)

This is a telescoping series. Most terms cancel out:

• \(+\sqrt{15}\) cancels with \(-\sqrt{15}\)
• \(-\sqrt{14}\) cancels with \(+\sqrt{14}\)
• \(+\sqrt{13}\) cancels with \(-\sqrt{13}\)
• \(-\sqrt{12}\) cancels with \(+\sqrt{12}\)
• \(+\sqrt{11}\) cancels with \(-\sqrt{11}\)
• \(-\sqrt{10}\) cancels with \(+\sqrt{10}\)
• \(+3\) cancels with \(-3\)

The terms that remain are:

\(4\) (from the first term) \(-2\sqrt{2}\) (from the last term)

So, the sum is \(4 - 2\sqrt 2\).

Final Answer

The value of the given expression is \(4 - 2\sqrt 2\).

Revision Table: Key Steps in Simplifying Radical Expressions

Additional Information: Understanding Conjugates and Rationalization

The technique used here, rationalizing the denominator, is crucial when working with expressions involving square roots in the denominator. The goal is to remove the radical from the denominator, making the expression easier to work with, especially for further calculations or comparisons. The conjugate of a binomial \(a + b\sqrt{c}\) is \(a - b\sqrt{c}\). When you multiply a binomial by its conjugate, the result is always a rational number (or an expression without radicals in the specific form used here, like \(a^2 - bc\)).

For example:

• The conjugate of \(3 + \sqrt{5}\) is \(3 - \sqrt{5}\). Their product is \((3+\sqrt{5})(3-\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4\).
• The conjugate of \(\sqrt{7} - \sqrt{2}\) is \(\sqrt{7} + \sqrt{2}\). Their product is \((\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2}) = (\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5\).

In our problem, each denominator resulted in 1 after multiplying by the conjugate, which significantly simplified the terms before summing them up. This specific structure led to a telescoping sum where intermediate terms canceled out, a common pattern in series problems after rationalization.