Evaluating Radical Expressions

The question asks us to find the value of a complex expression involving square roots:

\[\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}-\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}\]

We will break down the expression into three parts and evaluate each one separately before combining them.

Part 1: Evaluating the First Fraction

Let's evaluate the first part: \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\). To simplify this fraction, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{7}+\sqrt{5}\).

\[\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} = \frac{(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})} \times \frac{(\sqrt{7}+\sqrt{5})}{(\sqrt{7}+\sqrt{5})}\]

Using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\) for the denominator and the square of a sum formula \((a+b)^2 = a^2 + 2ab + b^2\) for the numerator, we get:

Numerator: \((\sqrt{7}+\sqrt{5})^2 = (\sqrt{7})^2 + 2(\sqrt{7})(\sqrt{5}) + (\sqrt{5})^2 = 7 + 2\sqrt{35} + 5 = 12 + 2\sqrt{35}\)

Denominator: \((\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) = (\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2\)

So, the first part is:

\[\frac{12 + 2\sqrt{35}}{2} = \frac{2(6 + \sqrt{35})}{2} = 6 + \sqrt{35}\]

Part 2: Evaluating the Second Fraction

Now let's look at the second part: \(\frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}-\sqrt{10}}\). We can notice a relationship between the numbers inside the square roots in this fraction and the first one. We can rewrite \(\sqrt{14}\) as \(\sqrt{2 \times 7} = \sqrt{2}\sqrt{7}\) and \(\sqrt{10}\) as \(\sqrt{2 \times 5} = \sqrt{2}\sqrt{5}\).

Substitute these into the expression:

\[\frac{\sqrt{2}\sqrt{7}+\sqrt{2}\sqrt{5}}{\sqrt{2}\sqrt{7}-\sqrt{2}\sqrt{5}}\]

Factor out \(\sqrt{2}\) from both the numerator and the denominator:

\[\frac{\sqrt{2}(\sqrt{7}+\sqrt{5})}{\sqrt{2}(\sqrt{7}-\sqrt{5})}\]

Since \(\sqrt{2} \neq 0\), we can cancel out the \(\sqrt{2}\) term:

\[\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\]

This is the exact same expression as Part 1. Therefore, the value of the second part is also \(6 + \sqrt{35}\).

Part 3: Evaluating the Third Fraction

The third part is \(\frac{\sqrt{10}}{\sqrt{5}}\). We can simplify this using the property \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\).

\[\frac{\sqrt{10}}{\sqrt{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}\]

Combining the Parts

The original expression is (Part 1) \(\div\) (Part 2) + (Part 3). Substituting the values we found:

\[(6 + \sqrt{35}) \div (6 + \sqrt{35}) + \sqrt{2}\]

Since \(6 + \sqrt{35}\) is a non-zero value, dividing it by itself gives 1.

\[1 + \sqrt{2}\]

This can also be written as \(\sqrt{2} + 1\).

Conclusion

The value of the given expression is \(\sqrt{2} + 1\).

The original expression becomes:

\((6 + \sqrt{35}) \div (6 + \sqrt{35}) + \sqrt{2}\)

\(1 + \sqrt{2}\)

Revision Table: Simplifying Radical Expressions

• To rationalize a denominator of the form \(\sqrt{a} \pm \sqrt{b}\), multiply by the conjugate \(\sqrt{a} \mp \sqrt{b}\).
• Use the difference of squares formula: \((a-b)(a+b) = a^2 - b^2\).
• Use the square of a sum/difference formula: \((a \pm b)^2 = a^2 \pm 2ab + b^2\).
• Simplify square roots: \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) and \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\).

Additional Information: Conjugates and Rationalization

The conjugate of a binomial \((a+b)\) is \((a-b)\), and vice versa. When dealing with expressions involving square roots in the denominator, like \(\frac{1}{\sqrt{a}-\sqrt{b}}\), multiplying the numerator and denominator by the conjugate of the denominator helps eliminate the square roots from the denominator. This process is called rationalization. It makes the expression easier to work with, especially when performing addition or subtraction. For example, \(\frac{1}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} = \frac{\sqrt{7}+\sqrt{5}}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{\sqrt{7}+\sqrt{5}}{7-5} = \frac{\sqrt{7}+\sqrt{5}}{2}\).

Understanding how to manipulate expressions with square roots and knowing common radical properties are essential skills for solving problems like this one.