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Chapter 7: Problem 105

Write each quotient in lowest terms. Assume that all variables represent positive real numbers. $$ \frac{16-4 \sqrt{8}}{12} $$

Short Answer

Expert verified
\( \frac{4 - 2 \text{sqrt}(2)}{3} \)

Step by step solution

01

Simplify the Numerator

Factor out the greatest common factor from the numerator. In this case, the greatest common factor of 16 and 4 is 4. This gives us: \ \[ 16 - 4\text{sqrt}(8) = 4(4 - \text{sqrt}(8)) \]
02

Simplify the Expression

Replace the simplified form of the numerator back in the quotient: \ \[ \frac{4(4 - \text{sqrt}(8))}{12} \]
03

Simplify Using Division

Divide the numerator and the denominator by their greatest common factor, which is 4 in this case: \ \[ \frac{4(4 - \text{sqrt}(8))}{12} = \frac{4 - \text{sqrt}(8)}{3} \]
04

Simplify the Square Root (if needed)

Note that \ \[ \text{sqrt}(8) = 2 \text{sqrt}(2) \]. So the expression becomes: \ \[ \frac{4 - 2 \text{sqrt}(2)}{3} \]

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is a key skill in algebra. The goal is to break a polynomial into simpler parts called factors. When you factor a polynomial, you look for the greatest common factor (GCF) first.
For example, in the expression \(16 - 4 \sqrt{8}\), the GCF of 16 and 4 is 4. Therefore, you factor out 4 from the numerator: \[16 - 4 \sqrt{8} = 4(4 - \sqrt{8}) \]
This makes the expression simpler and easier to divide.
Always check if the factored polynomial can be simplified further.
Factoring helps in breaking down complex algebraic expressions into simpler ones.
Simplifying Square Roots
Simplifying square roots is necessary when dealing with algebraic expressions. A square root is simplified when there are no perfect square factors other than 1 under the square root symbol.
In our problem, we had \( \sqrt{8} \). First, identify the largest perfect square factor of 8, which is 4. Then, express 8 as a product: \[ \sqrt{8} = \sqrt{4 \cdot 2} =\sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2} \]
Thus, \( \sqrt{8} \) simplifies to \( 2 \sqrt{2} \).
Simplifying square roots helps in making the final expression neater and usually more intuitive.
Lowest Terms in Fractions
Simplifying a fraction to its lowest terms means reducing it so that the numerator and the denominator have no common factors other than 1.
In the original expression, we have the fraction \(\frac{16 - 4 \sqrt{8}}{12}\).
After factoring and simplifying, this becomes \(\frac{4 - 2 \sqrt{2}}{3}\).
To ensure it is in the lowest terms, check for any common factors in the simplified expression.
Since there are no further common factors between the numerator \(4 - 2 \sqrt{2} \) and the denominator \(3 \), the fraction is now in its lowest terms.
Simplifying fractions makes them easier to work with and understand.

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Most popular questions from this chapter

Simplify each expression. Assume that all variables represent positive real numbers. $$ -5 x^{7 / 6}\left(x^{5 / 6}-x^{-1 / 6}\right) $$

Write each quotient in lowest terms. Assume that all variables represent positive real numbers. $$ \frac{3-3 \sqrt{5}}{3} $$

Rationalize each denominator. Assume that all variables represent positive real numbers and that no denominators are 0. $$ \frac{m-4}{\sqrt{m}+2} $$

Tom owns a condominium in a high rise building on the shore of Lake Michigan, and has a beautiful view of the lake from his window. He discovered that he can find the number of miles to the horizon by multiplying 1.224 by the square root of his eye level in feet from the ground. Use Tom's discovery to do the following. (a) Write a formula that could be used to calculate the distance \(d\) in miles to the horizon from a height \(h\) in feet from the ground. (b) Tom lives on the \(14^{\text {th }}\) floor, which is \(150 \mathrm{ft}\) above the ground. His eyes are \(6 \mathrm{ft}\) above his floor. Use the for- mula from part (a) to calculate the distance, to the nearest tenth of a mile, that Tom can see to the horizon from his condominium window.

The illumination \(I\), in foot-candles, produced by a light source is related to the distance \(d\), in feet, from the light source by the equation $$ d=\sqrt{\frac{k}{I}} $$ where \(k\) is a constant. If \(k=640,\) how far from the light source will the illumination be 2 foot-candles? Give the exact value, and then round to the nearest tenth of a foot.
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