Secondary Math 1 Module 8.2 Answer Key
Introduction
Secondary Math 1 Module 8.2 is a crucial part of the math curriculum for high school students. This module focuses on various mathematical concepts such as polynomials, factoring, and quadratic equations. In order to excel in this module, it is important to have access to the answer key, which provides solutions and explanations to the exercises and problems. In this article, we will provide you with the answer key for Secondary Math 1 Module 8.2, along with a detailed explanation of each problem.
Problem 1
The first problem in Secondary Math 1 Module 8.2 is about solving a quadratic equation. The equation is given as follows: 2x^2 + 5x - 3 = 0. To solve this equation, we can use either factoring or the quadratic formula. Let's use factoring to find the solution.
Solution: To factor the quadratic equation, we need to find two numbers whose product is equal to the product of the coefficients of x^2 and the constant term (in this case, -3), and whose sum is equal to the coefficient of x (in this case, 5).
By trial and error, we can find that the factors of -3 that add up to 5 are 3 and -1. Therefore, we can rewrite the equation as (2x + 3)(x - 1) = 0.
Setting each factor equal to zero, we get 2x + 3 = 0 and x - 1 = 0. Solving these equations, we find x = -3/2 and x = 1.
Therefore, the solutions to the quadratic equation 2x^2 + 5x - 3 = 0 are x = -3/2 and x = 1.
Problem 2
The second problem in Secondary Math 1 Module 8.2 deals with graphing quadratic functions. Given the equation f(x) = x^2 - 4x + 3, we are asked to graph this function on a coordinate plane.
Solution: To graph a quadratic function, we need to find its vertex and the direction of its opening. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c can be found using the formula x = -b/2a and y = f(x).
In this case, a = 1, b = -4, and c = 3. Plugging these values into the formula, we get x = 2 and y = -1. Therefore, the vertex of the function f(x) = x^2 - 4x + 3 is (2, -1).
Next, we need to determine the direction of the opening. Since the coefficient of x^2 is positive, the parabola opens upwards.
Using this information, we can now plot the vertex and draw the parabola. The graph should resemble a U-shape, opening upwards, with the vertex at (2, -1).
Problem 3
The third problem in Secondary Math 1 Module 8.2 involves solving a system of equations. The system is given as follows:
2x + y = 5
x - y = 1
Solution: To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From the second equation, we can isolate x as follows: x = y + 1.
Substituting this expression for x into the first equation, we get 2(y + 1) + y = 5. Simplifying the equation, we have 2y + 2 + y = 5, which further simplifies to 3y + 2 = 5.
Subtracting 2 from both sides of the equation, we get 3y = 3. Dividing both sides by 3, we find y = 1.
Substituting this value of y into the equation x = y + 1, we have x = 1 + 1, which gives us x = 2.
Therefore, the solution to the system of equations 2x + y = 5 and x - y = 1 is x = 2 and y = 1.
Problem 4
The fourth problem in Secondary Math 1 Module 8.2 requires us to find the zeros of a quadratic function. The function is given as f(x) = 4x^2 - 9x + 2.
Solution: To find the zeros of a quadratic function, we need to set the function equal to zero and solve for x. In this case, we have 4x^2 - 9x + 2 = 0.
Since factoring may not be straightforward in this case, we can use the quadratic formula to find the solutions. The quadratic formula states that for a quadratic equation ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac)) / 2a.
Applying this formula to our equation, we have x = (-(-9) ± √((-9)^2 - 4(4)(2))) / (2(4)).
Simplifying the equation, we get x = (9 ± √(81 - 32)) / 8.
Further simplifying, we have x = (9 ± √49) / 8, which gives us x = (9 ± 7) / 8.
Therefore, the solutions to the quadratic equation 4x^2 - 9x + 2 = 0 are x = 1 and x = 2/4.
Problem 5
The fifth problem in Secondary Math 1 Module 8.2 involves solving a word problem using quadratic equations. The problem is as follows:
A ball is thrown into the air with an initial velocity of 30 m/s. Its height in meters above the ground after t seconds is given by the equation h(t) = -5t^2 + 30t + 1. How long will it take for the ball to reach a height of 20 meters?
Solution: To solve this word problem, we need to find the value of t when h(t) = 20. In other words, we need to solve the equation -5t^2 + 30t + 1 = 20.
Using the quadratic formula, we have t = (-30 ± √(30^2 - 4(-5)(1-20))) / (2(-5)).
Simplifying the equation, we get t = (-30 ± √(900 + 400)) / (-10).
Further simplifying, we have t = (-30 ± √1300) / (-10), which gives us t = (-30 ± √(13 * 100)) / (-10).
Therefore, the solutions to the quadratic equation -5t^2 + 30t + 1 = 20 are t ≈ 0.18 and t ≈ 5.82.
Since time cannot be negative, the ball will reach a height of 20 meters after approximately 5.82 seconds.
Conclusion
Secondary Math 1 Module 8.2 covers various important topics in mathematics such as quadratic equations, factoring, and systems of equations. By having access to the answer key for this module, students can check their work and gain a better understanding of the concepts. In this article, we have provided the answer key for Module 8.2, along with detailed solutions and explanations for each problem. By using this answer key as a resource, students can improve their problem-solving skills and excel in Secondary Math 1.