Unit 8 Quadratic Equations Homework 10 Quadratic Word Problems Extra Quality

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How to Solve Quadratic Word Problems in Unit 8 Quadratic Equations Homework

Quadratic word problems are one of the most challenging topics in Unit 8 Quadratic Equations Homework 10. They involve applying the concepts of quadratic equations, such as factoring, completing the square, and using the quadratic formula, to real-world situations. In this article, we will show you how to solve quadratic word problems step by step, using examples and tips. By the end of this article, you will be able to master quadratic word problems and ace your homework.

What are Quadratic Word Problems?

Quadratic word problems are problems that can be modeled by quadratic equations. A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and a is not zero. A quadratic equation can have two solutions, called roots or zeros, which can be found by various methods.

Quadratic word problems often involve finding the maximum or minimum value of a quantity, such as area, profit, height, or speed. They can also involve finding the time or distance when two objects meet or collide, or finding the dimensions of a shape given its area or perimeter. Some common types of quadratic word problems are:

Area problems: These involve finding the length and width of a rectangle or a square given its area or perimeter.

Projectile problems: These involve finding the height, range, or time of flight of an object thrown or launched into the air.

Profit problems: These involve finding the price, revenue, cost, or profit of a product or service given a quadratic function.

Motion problems: These involve finding the speed, distance, or time of two objects moving towards or away from each other.

Pythagorean theorem problems: These involve finding the length of a side of a right triangle given the other two sides.

To solve quadratic word problems, you need to follow these steps:

Read and understand the problem carefully. Identify what is given and what is asked.

Write a quadratic equation that models the problem. Use variables to represent unknown quantities and constants to represent known quantities.

Solve the quadratic equation by your preferred method. You can use factoring, completing the square, or the quadratic formula. Check your solutions for validity and relevance.

Answer the question in words and units. Write your answer in a complete sentence and include appropriate units.

Let's look at some examples of quadratic word problems and how to solve them.

Example 1: Area Problem

The area of a rectangle is 120 square meters. The length of the rectangle is 6 meters more than twice its width. Find the dimensions of the rectangle.

Solution:

Step 1: Read and understand the problem. We are given the area of a rectangle and the relationship between its length and width. We are asked to find the dimensions of the rectangle.

Step 2: Write a quadratic equation that models the problem. Let x be the width of the rectangle. Then, the length of the rectangle is 2x + 6. The area of a rectangle is given by length times width, so we have:

A = lw

120 = (2x + 6)x

Expand and simplify:

120 = 2x^2 + 6x

0 = 2x^2 + 6x - 120

Divide by 2:

0 = x^2 + 3x - 60

This is our quadratic equation.

Step 3: Solve the quadratic equation by your preferred method. We can use factoring to solve this equation. We need to find two numbers that multiply to -60 and add to 3. These numbers are 10 and -6. So, we have:

x^2 + 3x - 60 = (x + 10)(x - 6)

Set each factor equal to zero and solve for x:

x + 10 = 0 or x - 6 = 0

x = -10 or x = 6

We have two possible solutions for x, but only one makes sense in the context of the problem. The width of a rectangle cannot be negative, so we reject x = -10. Therefore, x = 6 is our solution.

Step 4: Answer the question in words and units. We have found the width of the rectangle to be x = 6 meters. To find the length of the rectangle, we plug x into the expression for the length: l = 2x + 6 = 2(6) + 6 = 18 meters. So, our answer is:

The dimensions of the rectangle are 6 meters by 18 meters.

Example 2: Projectile Problem

A ball is thrown upward from the top of a building that is 80 meters high. The height of the ball in meters after t seconds is given by the function h(t) = -5t^2 + 20t + 80. Find the maximum height of the ball and the time it takes to reach that height.

Solution:

Step 1: Read and understand the problem. We are given a function that models the height of a ball thrown upward from a building. We are asked to find the maximum height of the ball and the time it takes to reach that height.

Step 2: Write a quadratic equation that models the problem. We already have a quadratic equation that models the problem. It is:

h(t) = -5t^2 + 20t + 80

This is our quadratic equation.

Step 3: Solve the quadratic equation by your preferred method. We can use completing the square to solve this equation. We need to make the coefficient of t^2 equal to 1, so we divide both sides by -5:

-h(t)/5 = t^2 - 4t - 16

Add 16 to both sides:

-h(t)/5 + 16 = t^2 - 4t

Complete the square by adding (b/2)^2 to both sides, where b is the coefficient of t:

-h(t)/5 + 16 + (-4/2)^2 = t^2 - 4t + (-4/2)^2

Simplify:

-h(t)/5 + 20 = (t - 2)^2

Subtract 20 from both sides:

-h(t)/5 = (t - 2)^2 - 20

Multiply both sides by -5:

h(t) = -5(t - 2)^2 + 100

This is our quadratic equation in vertex form.

Step 4: Answer the question in words and units. We can see from the vertex form of the equation that the maximum height of the ball is given by the constant term, which is 100 meters. The time it takes to reach that height is given by the value of t that makes (t - 2)^2 equal to zero, which is t = 2 seconds. So, our answer is:

The maximum height of the ball is 100 meters and it takes 2 seconds to reach that height.

Example 3: Profit Problem

A company sells x units of a product at a price of p dollars per unit. The revenue function is R(x) = px and the cost function is C(x) = 500 + 2x, where 500 is the fixed cost and 2 is the variable cost per unit. The profit function is P(x) = R(x) - C(x). Find the number of units that must be sold to break even and the number of units that must be sold to maximize the profit.

Solution:

Step 1: Read and understand the problem. We are given the revenue, cost, and profit functions of a company. We are asked to find the break-even point and the profit-maximizing point.

Step 2: Write a quadratic equation that models the problem. We already have a quadratic equation that models the problem. It is:

P(x) = R(x) - C(x)

Substitute the expressions for R(x) and C(x):

P(x) = px - (500 + 2x)

Expand and simplify:

P(x) = px - 500 - 2x

Rearrange the terms:

P(x) = -2x + px - 500

This is our quadratic equation.

Step 3: Solve the quadratic equation by your preferred method. We can use the quadratic formula to solve this equation. The quadratic formula is:

x = (-b ± √(b^2 - 4ac))/(2a)

where a, b, and c are the coefficients of x^2, x, and the constant term respectively.

In our equation, a = -2, b = p, and c = -500. So, we have:

x = (-(p) ± √((p)^2 - 4(-2)(-500)))/(2(-2))

Simplify:

x = (-p ± √(p^2 - 4000))/(-4)

We have two possible solutions for x, but only one makes sense in the context of the problem. The number of units sold cannot be negative, so we reject the solution with the negative sign. Therefore, x = (-p + √(p^2 - 4000))/(-4) is our solution.

Step 4: Answer the question in words and units. We have found the number of units that must be sold to break even. This is when the profit is zero, so we set P(x) equal to zero and solve for x:

0 = -2x + px - 500

Add 500 to both sides:

500 = -2x + px

Rearrange the terms:

-2x + px - 500 = 0

Factor out x:

x(-2 + p) = 500

Divide both sides by (-2 + p):

x = 500/(-2 + p)

This is the break-even point. The company must sell 500/(-2 + p) units to break even.

We have also found the number of units that must be sold to maximize the profit. This is when the vertex of the parabola is at its highest point. The x-coordinate of the vertex is given by -b/(2a), where b and a are the coefficients of x and x^2 respectively. In our equation, b = p and a = -2. So, we have:

x = -(p)/(2(-2))

Simplify:

x = p/4

This is the profit-maximizing point. The company must sell p/4 units to maximize the profit.

So, our answer is:

The company must sell 500/(-2 + p) units to break even and p/4 units to maximize the profit.

Conclusion

Quadratic word problems are problems that can be modeled by quadratic equations. They involve applying the concepts of quadratic equations, such as factoring, completing the square, and using the quadratic formula, to real-world situations. To solve quadratic word problems, you need to follow these steps:

Read and understand the problem carefully. Identify what is given and what is asked.

Write a quadratic equation that models the problem. Use variables to represent unknown quantities and constants to represent known quantities.

Solve the quadratic equation by your preferred method. You can use factoring, completing the square, or the quadratic formula. Check your solutions for validity and relevance.

Answer the question in words and units. Write your answer in a complete sentence and include appropriate units.

By following these steps, you can solve any quadratic word problem in Unit 8 Quadratic Equations Homework 10. You can also practice more problems and check your answers with online calculators or tutors. Quadratic word problems are not as hard as they seem. With some practice and patience, you can master them and ace your homework. 4aad9cdaf3