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Chapter 5: Problem 36

\(\frac{\sqrt{7}}{\sqrt{10}-\sqrt{2}}\)

Short Answer

Expert verified
\(\frac{\sqrt{70} + \sqrt{14}}{8}\)

Step by step solution

01

Rationalize the denominator

Multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{10} - \sqrt{2}\) is \(\sqrt{10} + \sqrt{2}\). \[\frac{\sqrt{7}}{\sqrt{10}-\sqrt{2}} \times \frac{\sqrt{10} + \sqrt{2}}{\sqrt{10} + \sqrt{2}}\]
02

Perform the multiplication

Now, perform the multiplication for both the numerator and the denominator. Remember, when you multiply two conjugates, the result is the difference of squares. For the numerator: \[\sqrt{7} \times (\sqrt{10} + \sqrt{2})\] And for the denominator: \[(\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2}) = (\sqrt{10})^2 - (\sqrt{2})^2 \]
03

Simplify the fraction

Simplify the fraction by applying the square roots, and perform additional operations if required, and simplify to the simplest form. \[\frac{\sqrt{70} + \sqrt{14}}{10 - 2}\] equals \[\frac{\sqrt{70} + \sqrt{14}}{8}\]

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Key Concepts

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

Quadratic Conjugates
Rationalizing the denominator involves using quadratic conjugates. This is a technique to eliminate radicals from the denominator of a fraction. Imagine a fraction like \( \frac{a}{\sqrt{b} - \sqrt{c}} \). Here, the conjugate of the denominator is \( \sqrt{b} + \sqrt{c} \). A conjugate just flips the sign between the two terms.

Multiplying the numerator and denominator by this conjugate does several things:
  • It removes the radical from the denominator.
  • Preserves the value of the original expression since multiplying by \( \frac{\sqrt{b} + \sqrt{c}}{\sqrt{b} + \sqrt{c}} \) is like multiplying by one.
A simple and efficient trick that makes more complex expressions manageable.
Simplifying Radical Expressions
Simplifying radical expressions is all about making expressions under radicals easier to interpret. Radicals, or square roots, can seem tricky, but simplifying them means reducing them to their simplest form so they are easy to work with.

Consider radicals like \( \sqrt{70} \). You can often simplify these by factoring out square numbers within the radical:
  • Look for square factors.
  • Extract these squares out of the radical.
This process involves understanding which numbers are perfect squares and which aren't, allowing you to break down and simplify the expression into smaller, more useful parts.
Difference of Squares
The difference of squares is a tangibly unique algebraic identity that is key to rationalizing denominators since it helps express products of conjugates. It is illustrated by the expression \( a^2 - b^2 \), which equals \( (a-b)(a+b) \). This pattern relates directly to
  • Expanding expressions with perfect square terms.
  • Removing radicals in expressions.
For example, when multiplying conjugates like \( (\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2}) \), instead of working through detailed algebra, the identity makes it straightforward, delivering \( 10 - 2 \) or \( 8 \). This simplifies calculations drastically and leads to more straightforward results.

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Most popular questions from this chapter

In Exercises 35鈥�46, determine whether the inverse of \(f\) is a function. Then find the inverse. $$ f(x)=x^3-1 $$

In Exercises 49鈥�52, determine whether the functions are inverses. $$ f(x)=\sqrt[5]{\frac{x+9}{5}}, g(x)=5 x^5-9 $$

\(f(x)=\sqrt[3]{3 x^2-x}\)

PROBLEM SOLVING For a drag race car with a total weight of 3500 pounds, the speed \(s\) (in miles per hour) at the end of a race can be modeled by \(s=14.8 \sqrt[3]{p}\), where \(p\) is the power (in horsepower). Graph the function. a. Determine the power of a 3500 -pound car that reaches a speed of 200 miles per hour. b. What is the average rate of change in speed as the power changes from 1000 horsepower to 1500 horsepower?

\(g(x)=\sqrt{x}-5\)
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