The quadratic formula is a powerful tool in algebra used to find the solutions of quadratic equations, which are equations of the form ax² + bx + c = 0. This formula is expressed as:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
where a, b, and c are constants. The solutions (or roots) of the quadratic equation can help in various real-life situations, from physics problems to financial calculations. Let’s explore three diverse examples that illustrate how to use the quadratic formula effectively.
Imagine you are launching a basketball from a height of 2 meters, and you want to determine when it will hit the ground. The height of the ball can be modeled by the equation:
\[ h(t) = -4.9t^2 + 6t + 2 \]
where h(t) is the height in meters and t is the time in seconds. To find out when the ball hits the ground, set h(t) to 0:
\[ -4.9t^2 + 6t + 2 = 0 \]
Here, a = -4.9, b = 6, and c = 2. Plugging these values into the quadratic formula gives:
\[ t = \frac{{-6 \pm \sqrt{{6^2 - 4(-4.9)(2)}}}}{{2(-4.9)}} \]
This simplifies to:
\[ t = \frac{{-6 \pm \sqrt{{36 + 39.2}}}}{{-9.8}} \]
\[ t = \frac{{-6 \pm \sqrt{{75.2}}}}{{-9.8}} \]
Calculating the square root and simplifying yields two possible times. The positive value will give you the time when the basketball hits the ground.
Notes: This example illustrates how the quadratic formula helps in modeling real-world scenarios, such as projectile motion in sports.
Suppose you own a small business selling handmade crafts. Your profit, P, in dollars, can be represented by the equation:
\[ P(x) = -2x^2 + 40x - 100 \]
where x is the number of items sold. To find out how many items you need to sell to maximize your profit, you can set the equation to zero and find the vertex:
\[ -2x^2 + 40x - 100 = 0 \]
Here, a = -2, b = 40, and c = -100. Using the quadratic formula:
\[ x = \frac{{-40 \pm \sqrt{{40^2 - 4(-2)(-100)}}}}{{2(-2)}} \]
This simplifies to:
\[ x = \frac{{-40 \pm \sqrt{{1600 - 800}}}}{{-4}} \]
\[ x = \frac{{-40 \pm \sqrt{{800}}}}{{-4}} \]
Calculating the values will provide you with two potential sales quantities, from which you can determine the maximum profit.
Notes: This example shows how businesses can apply the quadratic formula to optimize their operations and profits.
Let’s say you want to design a rectangular garden. The area A of the garden can be expressed as:
\[ A = x(x + 5) \]
where x is the width in meters, and the length is 5 meters more than the width. If you want the area to be 60 square meters, set up the equation:
\[ x(x + 5) - 60 = 0 \]
This expands to:
\[ x^2 + 5x - 60 = 0 \]
Here, a = 1, b = 5, and c = -60. Plugging these into the quadratic formula yields:
\[ x = \frac{{-5 \pm \sqrt{{5^2 - 4(1)(-60)}}}}{{2(1)}} \]
After simplifying:
\[ x = \frac{{-5 \pm \sqrt{{25 + 240}}}}{{2}} \]
\[ x = \frac{{-5 \pm \sqrt{{265}}}}{{2}} \]
Determine the positive root to find the width of the garden, and then calculate the length.
Notes: This example highlights how homeowners can use the quadratic formula for practical planning in gardening and landscaping.