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Chapter 10: Problem 68

Rationalize each denominator. $$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{5}+\sqrt{2}}$$

Short Answer

Expert verified
The rationalized form is \[ \frac{\sqrt{35} - \sqrt{14} + 5 - \sqrt{10}}{3} \].

Step by step solution

01

Identify the conjugate

To rationalize the denominator, we first need to identify the conjugate of the denominator. The conjugate of \(\sqrt{5} + \sqrt{2}\) is \(\sqrt{5} - \sqrt{2}\).
02

Multiply numerator and denominator by the conjugate

Multiply both the numerator and the denominator by the conjugate of the denominator: \[\frac{\sqrt{7} + \sqrt{5}}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}\].
03

Apply the distributive property (FOIL method) in the numerator

Expand the numerator using the distributive property: \[ (\sqrt{7} + \sqrt{5})(\sqrt{5} - \sqrt{2}) = (\sqrt{7} \cdot \sqrt{5}) + (\sqrt{7} \cdot -\sqrt{2}) + (\sqrt{5} \cdot \sqrt{5}) + (\sqrt{5} \cdot -\sqrt{2}) \]. Simplifying this gives us: \[\sqrt{35} - \sqrt{14} + 5 - \sqrt{10}\].
04

Apply the difference of squares in the denominator

Expand and simplify the denominator using the difference of squares formula \((a+b)(a-b) = a^2 - b^2\). Here \(a = \sqrt{5}\) and \(b = \sqrt{2}\), so \[ (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = \sqrt{5}^2 - \sqrt{2}^2 = 5 - 2 = 3 \].
05

Combine and simplify the result

Place the simplified numerator over the simplified denominator: \[ \frac{\sqrt{35} - \sqrt{14} + 5 - \sqrt{10}}{3}\]. This is the rationalized form of the given expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conjugate
To understand how to rationalize the denominator, knowing about the 'conjugate' is key. A conjugate is found by changing the sign between two terms in a binomial. For example, the conjugate of \( \sqrt{5} + \sqrt{2} \) is \( \sqrt{5} - \sqrt{2} \). When we multiply a binomial by its conjugate, we use the formula \( (a+b)(a-b) = a^2 - b^2 \). This procedure helps in eliminating the radicals from the denominator
Distributive Property
When rationalizing expressions, you often use the distributive property to simplify. This property states that \(a(b + c) = ab + ac\). In the given exercise, using the distributive property helps expand and simplify the numerator. For example, multiplying out \( (\sqrt{7} + \sqrt{5})(\sqrt{5} - \sqrt{2}) \) gives \(\sqrt{7} \cdot\sqrt{5} + \sqrt{7} \cdot -\sqrt{2} + \sqrt{5} \cdot\sqrt{5} + \sqrt{5} \cdot -\sqrt{2} \). By applying this property, the numerator simplifies to \(\sqrt{35} - \sqrt{14} + 5 - \sqrt{10} \).
Difference of Squares
The difference of squares is a powerful technique often used to rationalize denominators. It states that \( (a+b)(a-b) = a^2 - b^2 \). This formula reduces a binomial product into the difference of squares of the terms. Applying this to our denominator \( \sqrt{5} + \sqrt{2} \) and its conjugate \( \sqrt{5} - \sqrt{2} \), we get \( \sqrt{5} \cdot\sqrt{5} - \sqrt{2} \cdot\sqrt{2} \) which simplifies to \ 5 - 2 = 3 \. This makes the denominator a rational number, removing the radicals.

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