Problem 88
Question
Complete the statement with \(<,=\), or \(\geq$$\sqrt{\frac{3}{11}} \quad \frac{\sqrt{3}}{\sqrt{11}}\)
Step-by-Step Solution
Verified Answer
\( \sqrt{\frac{3}{11}} = \frac{\sqrt{3}}{\sqrt{11}} \)
1Step 1: Analyse the square root property
The property in focus is: \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \). This statement shows that the square root of a fraction is equal to the fraction of the square roots.
2Step 2: Simplify \( \sqrt{\frac{3}{11}} \)
Following the earlier identified property, we rewrite \( \sqrt{\frac{3}{11}} = \frac{\sqrt{3}}{\sqrt{11}} \)
3Step 3: Compare the two expressions
It has been established that \( \sqrt{\frac{3}{11}} \) = \( \frac{\sqrt{3}}{\sqrt{11}} \). Thus, these two expressions are equal.
Key Concepts
Properties of Square RootsSimplifying Square RootsComparison of Square Root Expressions
Properties of Square Roots
Understanding the properties of square roots can greatly simplify handling complex expressions. One essential property is that the square root of a fraction can be separated into the roots of its numerator and denominator. This is expressed as \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \). This property is very useful because it allows us to break down complicated square root expressions into simpler components.
Take the example of \( \sqrt{\frac{3}{11}} \). By applying this property, we can rework the expression as \( \frac{\sqrt{3}}{\sqrt{11}} \). This tells us that the values are indeed equivalent. Remember, these properties hold true only if both 'a' and 'b' are non-negative, as square roots of negative numbers introduce complex values. This property can make solving problems more manageable and less intimidating.
Take the example of \( \sqrt{\frac{3}{11}} \). By applying this property, we can rework the expression as \( \frac{\sqrt{3}}{\sqrt{11}} \). This tells us that the values are indeed equivalent. Remember, these properties hold true only if both 'a' and 'b' are non-negative, as square roots of negative numbers introduce complex values. This property can make solving problems more manageable and less intimidating.
Simplifying Square Roots
Simplifying square roots is a powerful mathematical tool to reduce complexity. To simplify square roots of fractions, we often employ the previously discussed property, \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \). Such simplification helps us to see more clearly how components of an expression relate to one another.
In some cases, simplifying square roots not only involves separating the numerator and denominator but also breaking them down into their prime factors. For example, if we have \( \sqrt{50} \), we can express this as \( \sqrt{25 \times 2} \), which can be further simplified to \( 5\sqrt{2} \).
In some cases, simplifying square roots not only involves separating the numerator and denominator but also breaking them down into their prime factors. For example, if we have \( \sqrt{50} \), we can express this as \( \sqrt{25 \times 2} \), which can be further simplified to \( 5\sqrt{2} \).
- Always look for perfect squares within the expression.
- Use the commutative property of multiplication (order doesn't matter in multiplication), to rearrange terms conveniently.
Comparison of Square Root Expressions
When comparing square root expressions, it is important to have a consistent basis for comparison. Very often, transforming expressions using known properties can reveal equivalency or inequality. For instance, by applying the property \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \), both expressions like \( \sqrt{\frac{3}{11}} \) and \( \frac{\sqrt{3}}{\sqrt{11}} \) become more uniform and comparable directly.
Comparison can involve:
Comparison can involve:
- Checking for direct equivalency as shown in the initial practice.
- Comparing simplified forms.
- Using numerical approximations when exact simplification seems uneasily accessible.
Other exercises in this chapter
Problem 86
Complete the statement with \(
View solution Problem 87
Complete the statement with \(
View solution Problem 93
The escape velocity (in meters per second) on the moon is \(\sqrt{\frac{2\left(6.67 \times 10^{-11}\right)\left(7.36 \times 10^{22}\right)}{1.74 \times 10^{6}}}
View solution Problem 94
The escape velocity (in meters per second) on Mars is \(\sqrt{\frac{2\left(6.67 \times 10^{-11}\right)\left(6.42 \times 10^{23}\right)}{3.37 \times 10^{6}}}\) W
View solution