Problem 103 - College Algebra

Question

The algebraic expression \(2 \sqrt{5 L}\) is used to estimate the speed of a car prior to an accident, in miles per hour, based on the length of its skid marks, \(L,\) in feet. Find the speed of a car that left skid marks 40 feet long, and write the answer in simplified radical form.

Step-by-Step Solution

Answer

The speed of a car that left skid marks that are 40 feet long is estimated to be \(4\sqrt{50}\) miles per hour.

Step 1: Substitute the Given Length into the Algebraic Expression

Substitute \(L= 40\) feet into the algebraic expression \(2 \sqrt{5 L}\) to get: \[2 \sqrt{5 \times 40}\]

Step 2: Simplify the Expression within the Square Root

Carry out the operation within the square root to obtain: \[2 \sqrt{200}\]

Step 3: Simplify the Square Root

Simplify \(\sqrt{200}\) to \(\sqrt{4}\times\sqrt{50}\) which further simplifies to \(2\sqrt{50}\) due to the property \(\sqrt{ab} = \sqrt{a}\times\sqrt{b}\). So the expression now is: \[2 \times 2\sqrt{50}\]

Step 4: Simplify the Overall Expression

Finally, multiply the numbers outside the square root to find the speed: \[4\sqrt{50}\] miles per hour

Key Concepts

Simplified Radical Form

Understanding the simplified radical form is crucial when dealing with algebraic expressions involving square roots. In this context, a radical is an expression that includes a square root. To simplify it, you aim to find the largest perfect square factor of the number under the square root and rewrite the radical as a product of two square roots, one of which is a perfect square. This helps in reducing the expression to its simplest form.
For example, when simplifying \(\sqrt{200}\), you first identify 100 as a perfect square factor (as \(100 \times 2 = 200\)). This allows you to rewrite \(\sqrt{200}\) as \(\sqrt{100} \times \sqrt{2}\). Since \(\sqrt{100}\) equals 10, it further simplifies to \(10\sqrt{2}\).
This simplification method helps not only in clearer calculations but also in providing clearer results, such as in algebraic expressions like the one mentioned in the exercise. Simplified radicals provide results that are easier to work with, both numerically and algebraically.

Square Root Simplification

Square root simplification involves breaking down a radical into its simplest form by eliminating perfect square factors under the root. This is essential for making calculations manageable and ensuring that expressions are as compact as possible.
To simplify a square root like \(\sqrt{200}\), first look for perfect square factors. In 200, you find 4 and 50, since \(4 \times 50 = 200\). Simplifying, you can write \(\sqrt{200}\) as \(\sqrt{4 \times 50}\).

• Extract the \(\sqrt{4}\) as 2, leading to \(2\sqrt{50}\).
• If possible, continue simplifying \(\sqrt{50}\) as \(\sqrt{25 \times 2}\), bringing it further to \(5\sqrt{2}\) because \(\sqrt{25} = 5\).
• Combine to get \(2 \times 5\sqrt{2} = 10\sqrt{2}\).

By understanding and practicing this method, you can simplify any square root more comfortably, allowing you to focus on solving broader mathematical problems.

Solving Word Problems

Word problems are puzzles in mathematics where you derive equations from real-world scenarios and solve them using algebraic principles. In word problems, like the one in the exercise, the challenge is to interpret the given information, translate it into mathematical expressions, and then solve it step by step.
Here's how to tackle them effectively:

• Read the problem carefully to understand what is being asked.
• Identify and note down known quantities and variables. In this case, \(L = 40\) feet of skid marks.
• Convert the word problem into an algebraic expression, as seen with the expression \(2 \sqrt{5L}\).
• Simplify the algebraic expression using mathematical operations, including substituting known values and simplifying radicals.

Through practice, solving word problems can become more intuitive and less daunting. You start to recognize patterns and develop strategies that efficiently lead from linguistic descriptions to numeric solutions.