# 50 Lesson 9.2 Algebra 1 Answers

## Introduction

Algebra 1 is a fundamental course that introduces students to the world of algebraic equations and problem-solving. Lesson 9.2 in Algebra 1 covers a variety of topics, including solving quadratic equations. In this article, we will provide detailed answers and explanations for the problems in Lesson 9.2, helping students gain a better understanding of the subject matter.

## Solving Quadratic Equations

### Problem 1: Solve the quadratic equation x^2 + 5x + 6 = 0

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). In this case, a = 1, b = 5, and c = 6. Plugging these values into the formula, we get x = (-5 ± √(5^2 - 4(1)(6))) / (2(1)). Simplifying further, x = (-5 ± √(25 - 24)) / 2, which gives us two solutions: x = (-5 + 1) / 2 and x = (-5 - 1) / 2. Therefore, the solutions to the quadratic equation are x = -2 and x = -3.

### Problem 2: Solve the quadratic equation 2x^2 - 7x - 3 = 0

Using the quadratic formula again, we can find the solutions to this equation. In this case, a = 2, b = -7, and c = -3. Substituting these values into the quadratic formula, we get x = (-(-7) ± √((-7)^2 - 4(2)(-3))) / (2(2)). Simplifying further, x = (7 ± √(49 + 24)) / 4, which gives us two solutions: x = (7 + 5) / 4 and x = (7 - 5) / 4. Therefore, the solutions to the quadratic equation are x = 3 and x = -0.5.

## Factoring Quadratic Equations

### Problem 3: Factor the quadratic equation x^2 + 3x + 2

To factor this quadratic equation, we need to find two numbers whose sum is 3 and whose product is 2. In this case, the numbers are 1 and 2. Therefore, we can rewrite the equation as (x + 1)(x + 2) = 0. Setting each factor equal to zero, we get x + 1 = 0 and x + 2 = 0. Solving for x, we find that the solutions to the equation are x = -1 and x = -2.

### Problem 4: Factor the quadratic equation 2x^2 + 5x + 3

Similarly, to factor this quadratic equation, we need to find two numbers whose sum is 5 and whose product is 3. In this case, the numbers are 2 and 3. Thus, we can rewrite the equation as (2x + 3)(x + 1) = 0. Setting each factor equal to zero, we obtain 2x + 3 = 0 and x + 1 = 0. Solving for x, we find that the solutions to the equation are x = -1 and x = -1.5.

## Completing the Square

### Problem 5: Solve the quadratic equation x^2 + 8x + 16 = 0 by completing the square

To solve this quadratic equation by completing the square, we need to rewrite it in the form (x + a)^2 = 0. In this case, we have (x + 4)^2 = 0. Taking the square root of both sides, we get x + 4 = 0. Solving for x, we find that the solution to the equation is x = -4.

### Problem 6: Solve the quadratic equation 3x^2 + 6x + 2 = 0 by completing the square

Similarly, to solve this quadratic equation by completing the square, we need to rewrite it as (x + a)^2 = 0. In this case, we have (x + 1)^2 = -1. Taking the square root of both sides, we get x + 1 = ±√(-1). However, since the square root of a negative number is not a real number, this equation has no real solutions.

## Using the Quadratic Formula

### Problem 7: Solve the quadratic equation 4x^2 + 3x - 5 = 0 using the quadratic formula

Using the quadratic formula, we can find the solutions to this equation. In this case, a = 4, b = 3, and c = -5. Substituting these values into the formula, we get x = (-3 ± √(3^2 - 4(4)(-5))) / (2(4)). Simplifying further, x = (-3 ± √(9 + 80)) / 8, which gives us two solutions: x = (-3 + √89) / 8 and x = (-3 - √89) / 8.

### Problem 8: Solve the quadratic equation 2x^2 - 8x + 4 = 0 using the quadratic formula

Applying the quadratic formula to this equation, with a = 2, b = -8, and c = 4, we get x = (-(-8) ± √((-8)^2 - 4(2)(4))) / (2(2)). Simplifying further, x = (8 ± √(64 - 32)) / 4, which gives us two solutions: x = (8 + √32) / 4 and x = (8 - √32) / 4.

## Conclusion

Lesson 9.2 in Algebra 1 covers the various methods of solving quadratic equations, including using the quadratic formula, factoring, and completing the square. By understanding and practicing these methods, students can become proficient in solving quadratic equations and further their mathematical skills. We hope that this article has provided helpful answers and explanations for the problems in Lesson 9.2, allowing students to improve their understanding of algebraic concepts.