91Ó°ÊÓ

A quadratic equation is an essential part of algebra that takes the form \( ax^2 + bx + c = 0 \). These equations feature a variable (usually represented as 'x') that is squared, meaning the highest exponent of the variable is 2. Understanding quadratic equations is key to solving many real-world problems, like the one Fiona faces with her boat.

In Fiona's problem, we arrive at the quadratic equation \[ 4x^2 - 45x - 36 = 0 \] when calculating the boat speed. To solve quadratic equations, you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where 'a', 'b', and 'c' are coefficients from the quadratic equation.

Plugging in our values, we find two solutions: 12 and -0.75. Since a boat's speed can't be negative, our valid solution is 12, meaning the speed of Fiona's boat in still water is 12 mph.

One of the fundamental concepts in physics and algebra is the relationship between time, distance, and speed. The formula connecting these three aspects is: \[ \text{time} = \frac{\text{distance}}{\text{speed}} \].

Using Fiona's problem, let's break it down:

• Going upstream, Fiona's effective speed is reduced due to the river current, resulting in \( x - 3 \) mph.
• Going downstream, the current aids her boat, making the effective speed \( x + 3 \) mph.

The time for each leg of Fiona's trip is:

• Upstream: \( \frac{45}{x - 3} \) hours
• Downstream: \( \frac{45}{x + 3} \) hours

Combining these times gives us the total trip time of 8 hours, leading to the equation: \[ \frac{45}{x - 3} + \frac{45}{x + 3} = 8 \].
Understanding and applying this simple relationship helps solve many similar real-life problems.

Solving equations systematically is a vital skill in both mathematics and everyday problem-solving. Here’s a step-by-step approach using Fiona's problem as an example:

• **Identify the unknown variable:** In this case, the speed of the boat in still water is represented as \( x \).
• **Set up the equation:** Based on the total travel time and given distances, we form the equation \[ \frac{45}{x - 3} + \frac{45}{x + 3} = 8 \].
• **Simplify the equation:** Combine terms by finding a common denominator, in this case, \( (x - 3)(x + 3) \), leading to \[ \frac{45(x + 3) + 45(x - 3)}{(x - 3)(x + 3)} = 8 \].
• **Solve for the variable:** This simplifies further to \[ 90x = 8x^2 - 72 \], and eventually to the quadratic equation \[ 4x^2 - 45x - 36 = 0 \].
  Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solution. For our coefficients \( a = 4 \), \( b = -45 \), and \( c = -36 \), we find \( x = 12 \) (valid speed) and \( x = -0.75 \) (not valid as speed can't be negative).

These steps illustrate a systematic approach to solving equations, reinforcing logical problem-solving skills.