Solving Quadratic Word Problems In Algebra 1 Homework

[Jocelin Taylor]

Solving Quadratic Word Problems in Algebra 1 Homework

Quadratic word problems are problems that involve quadratic equations or functions. A quadratic equation is an equation of the form ax + bx + c = 0, where a, b, and c are constants and x is a variable. A quadratic function is a function of the form f(x) = ax + bx + c, where a, b, and c are constants and x is a variable.

solving quadratic word problems in algebra 1 homework

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Quadratic word problems often involve situations that can be modeled by quadratic functions, such as projectile motion, area, volume, profit, etc. To solve quadratic word problems, we need to follow these steps:

• Read the problem carefully and identify the given information and the unknown quantity.

• Write a quadratic equation or function that represents the problem. If necessary, use the quadratic formula or factoring to find the solutions of the equation.

• Check if the solutions make sense in the context of the problem and answer the question.

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

Example 1: Projectile Motion

A ball is thrown upward from the ground with an initial velocity of 15 meters per second. The height of the ball in meters after t seconds is given by the function h(t) = -5t + 15t. How long will it take for the ball to reach its maximum height? What is the maximum height?

To solve this problem, we need to find the vertex of the parabola defined by the function h(t). The vertex represents the maximum or minimum point of the parabola. The x-coordinate of the vertex is given by the formula -b/2a, where a and b are the coefficients of the quadratic term and the linear term, respectively. In this case, we have a = -5 and b = 15, so we get:

t=-b2a=-152(-5)=1510=1.5

This means that it will take 1.5 seconds for the ball to reach its maximum height. To find the maximum height, we need to plug in this value into the function h(t):

h(1.5)=-51.52+15(1.5)=-11.25+22.5=11.25

The maximum height is 11.25 meters.

Example 2: Area of a Rectangle

The length of a rectangle is 3 meters more than its width. The area of the rectangle is 84 square meters. What are the dimensions of the rectangle?

To solve this problem, we need to write a quadratic equation that represents the area of the rectangle. Let x be the width of the rectangle. Then the length of the rectangle is x + 3. The area of the rectangle is the product of the length and the width, so we have:

A=lw=(x+3)x=84

Expanding and rearranging, we get:

x2+3x-84=0

This is a quadratic equation that we can solve by factoring. We need to find two numbers that multiply to -84 and add to 3. These numbers are 12 and -7, so we have:

x2+3x-84=0

```html mo>-7x+12=0

Using the zero product property, we get:

x-7=0 or x+12=0

Solving for x, we get:

x=7 or x=-12

Since the width of the rectangle cannot be negative, we reject the solution x = -12. Therefore, the width of the rectangle is x = 7 meters and the length of the rectangle is x + 3 = 10 meters.

Example 3: Profit of a Business

A business sells x units of a product at a price of $50 - $0.5x per unit. The cost of producing x units is $100 + $20x. What is the maximum profit that the business can make? How many units should the business sell to achieve this profit?

To solve this problem, we need to write a quadratic function that represents the profit of the business. The profit is the difference between the revenue and the cost, so we have:

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

The revenue is the product of the price and the quantity, so we have:

```html mo>)=(50-0.5x)x

The cost is given by the expression $100 + $20x, so we have:

C(x)=100+20x

Substituting these expressions into the profit function, we get:

P(x)=(50-0.5x)x-100-20x

Simplifying and rearranging, we get:

```html mo>)=-0.5x2-20.5x+2500

This is a quadratic function that has a negative leading coefficient, which means that it has a maximum value. To find the maximum value, we need to find the vertex of the parabola defined by the function P(x). The x-coordinate of the vertex is given by the formula -b/2a, where a and b are the coefficients of the quadratic term and the linear term, respectively. In this case, we have a = -0.5 and b = -20.5, so we get:

x=-b2a=-(-20.5)2(-0.5)=20.51=20.5

This means that the maximum profit occurs when the business sells 20.5 units of the product. To find the maximum profit, we need to plug in this value into the function P(x):

```html 20.5)=-0.520.52-20.5(20.5)+2500=-210.25-420.25+2500=1869.5

The maximum profit is $1869.5.

In conclusion, we have learned how to solve quadratic word problems in algebra 1 homework by following these steps:

• Read the problem carefully and identify the given information and the unknown quantity.

• Write a quadratic equation or function that represents the problem. If necessary, use the quadratic formula or factoring to find the solutions of the equation.

• Check if the solutions make sense in the context of the problem and answer the question.

We have also seen some examples of quadratic word problems involving projectile motion, area of a rectangle, and profit of a business. We hope this article has helped you understand how to solve quadratic word problems in algebra 1 homework.

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