91Ó°ÊÓ

When working with direct variation, the constant of variation plays a crucial role. It is the factor that links two directly varying quantities. With direct variation, one quantity increases or decreases in direct proportion to the square root or some other function of another quantity. In our problem, we express direct variation as:

• \( r = k \sqrt{n} \), where \( r \) varies directly as the square root of \( n \).

Here, \( k \) is the constant of variation. It remains the same for any values of \( r \) and \( n \) as long as the relationship is preserved. To find \( k \), you need at least one pair of specific values of \( r \) and \( n \). This known pair lets you solve for \( k \) using the equation:

• \( k = \frac{r}{\sqrt{n}} \)

Once you have the constant of variation, you can predict an unknown value in a direct variation problem with any other value.

Understanding the square root is vital in this context since our relationship involves \( \sqrt{n} \). The square root of a number \( n \) is a value that, when multiplied by itself, gives \( n \). It is written as \( \sqrt{n} \). This unique mathematical function simplifies many physics and geometry problems. Here are some key points about square roots:

• \( \sqrt{250} \) is an expression for which we will find a simpler form by factorizing to simplify calculations.
• The square root of perfect squares, like \( 160 \) being simplified to \( 4 \sqrt{10} \), is simpler and often more intuitive.

The square root manipulation often leads into more practical and comprehendible results in calculations, making predictions more straightforward.

In mathematics, solving equations is about finding the unknown variable that satisfies a mathematical statement. This requires actions like substitution and arithmetic simplification. Here's a breakdown of how we solve an equation in this scenario:

• Identify known and unknown quantities.
• Use equations that express a relationship. For example, \( r = k \sqrt{n} \).
• Substitute known values to find other variables (e.g., calculate \( k \)).

The essence of solving equations lies in isolating the unknown and processing each side of the equation equally, except if simplifying involves lessening the difficulty of expressions through cancellation or common factors.

Rationalizing the denominator is a valuable algebraic technique that involves eliminating any irrational numbers, such as square roots, from the bottom (denominator) of a fraction. In our example, we started with:

• \( \frac{4}{5\sqrt{10}} \)

To rationalize:

• Multiply both the numerator and denominator by \( \sqrt{10} \).
• This simplifies to \( \frac{4\sqrt{10}}{50} \).
• Ultimately, you obtain \( \frac{2\sqrt{10}}{25} \).

Rationalizing allows easier computational manipulation and a cleaner form of the expression. This simplification often aids further calculations, as seen while determining the value of \( r \).