Summary
Highlights
The Pythagorean theorem (a² + b² = c²) is introduced for right triangles, explaining the legs (a, b) and the hypotenuse (c).
An example demonstrates using the Pythagorean theorem to find a missing side 'a' when 'b' is 12 and 'c' is 15. The solution involves solving a simple quadratic equation, yielding a = 9, with emphasis on using only the positive root for length.
The video introduces the application of quadratic equations in various fields, starting with geometry and trigonometry, specifically the Pythagorean theorem.
A more complex application of the Pythagorean theorem is presented: a boat being pulled into a dock. The problem involves defining variables for the height of the dock (x) and the rope length (2x + 3) to form a right triangle.
The setup leads to the quadratic equation x² + 12² = (2x + 3)². The algebra is worked through, including expanding the binomial squared and simplifying the equation. The equation simplifies to 3x² + 12x - 135 = 0, which is then divided by 3 to get x² + 4x - 45 = 0. Factoring results in (x + 9)(x - 5) = 0, giving potential solutions of x = -9 and x = 5. The negative solution is discarded as length cannot be negative, so the height of the dock is 5 feet.
Another application involves calculating the dimensions of a rug within a room, leaving a uniform strip of floor around it. The room is 12 feet wide and 15 feet long, and the rug's area is 108 square feet. The uniform strip width is denoted as 'x'.
The rug's dimensions are expressed as (15 - 2x) by (12 - 2x). Setting the product of these dimensions equal to the rug's area (108) yields a quadratic equation: (15 - 2x)(12 - 2x) = 108. After expanding and simplifying, the equation becomes 4x² - 54x + 72 = 0. Dividing by 2 simplifies it to 2x² - 27x + 36 = 0. Factoring this equation leads to (2x - 3)(x - 12) = 0, giving solutions x = 3/2 and x = 12. The solution x = 12 is disregarded because it would make the width of the rug zero or negative. Using x = 3/2, the rug dimensions are calculated as 12 feet by 9 feet.
The final application is from physics, involving a projectile launched from ground level with an initial velocity of 96 feet per second. The height function is given as s = -16t² + 96t. The goal is to find the time(s) when the projectile reaches a height of 80 feet.
Setting the height function equal to 80 ( -16t² + 96t = 80) and rearranging gives a quadratic equation: -16t² + 96t - 80 = 0. Dividing by -16 simplifies it to t² - 6t + 5 = 0. Factoring results in (t - 5)(t - 1) = 0, yielding two times: t = 1 second and t = 5 seconds. This indicates the projectile is at 80 feet during both its ascent and descent.