How Solve This Task From Russian State Exam?

Introduction

The Russian State Exam is a challenging and comprehensive exam that tests a student's knowledge and skills in various subjects, including mathematics. One of the key areas of mathematics that is often tested in the exam is algebra, particularly equations with absolute values. In this article, we will focus on solving a specific type of equation that involves absolute values, and provide a step-by-step guide on how to approach it.

Understanding the Equation

The given equation is:

(|x− a− 2|+ |x− a+ 2|)2 − a(|x− a− 2|+ |x− a+ 2|) + a2 − 64 = 0

This equation involves absolute values, which can be challenging to work with. However, with a clear understanding of the concept of absolute values and how to handle them, we can break down the equation and solve it step by step.

Breaking Down the Equation

To solve the equation, we need to break it down into smaller parts and analyze each part separately. Let's start by looking at the absolute value expressions:

|x− a− 2| and |x− a+ 2|

These expressions represent the distance between x and a-2, and x and a+2, respectively. The absolute value of a quantity is its distance from zero, without considering whether it's positive or negative.

Case 1: x < a-2

When x < a-2, the absolute value expressions become:

|x− a− 2| = a-2 - x |x− a+ 2| = -(x-a+2) = a-2-x

Substituting these expressions into the original equation, we get:

((a-2-x) + (a-2-x))^2 - a((a-2-x) + (a-2-x)) + a^2 - 64 = 0

Simplifying the equation, we get:

(2a-4-2x)^2 - a(2a-4-2x) + a^2 - 64 = 0

Case 2: a-2 ≤ x < a+2

When a-2 ≤ x < a+2, the absolute value expressions become:

|x− a− 2| = x-a-2 |x− a+ 2| = a+2-x

Substituting these expressions into the original equation, we get:

((x-a-2) + (a+2-x))^2 - a((x-a-2) + (a+2-x)) + a^2 - 64 = 0

Simplifying the equation, we get:

(2)^2 - a(2) + a^2 - 64 = 0

Case 3: x ≥ a+2

When x ≥ a+2, the absolute value expressions become:

|x− a− 2| = x-a-2 |x− a+ 2| = x-a+2

Substituting these expressions into the original equation, we get:

((x-a-2) + (x-a+2))^2 - a((x-a-2) + (x-a+2)) + a^2 - 64 = 0

Simplifying the equation, we get:

(2x-2a)^2 - a(2x-2a) + a^2 - 64 = 0

Solving Equation

Now that we have broken down the equation into three cases, we can solve each case separately.

Case 1: x < a-2

Simplifying the equation, we get:

(2a-4-2x)^2 - a(2a-4-2x) + a^2 - 64 = 0

Expanding the squared term, we get:

4a^2-16a+16+4x2-16ax+16-2a(2a-4-2x) + a^2 - 64 = 0

Simplifying the equation, we get:

4a^2-16a+16+4x2-16ax-4a^{2+8ax+2ax-4x+a}2-64 = 0

Combine like terms:

4x^2-12ax+4x+a2-48 = 0

Case 2: a-2 ≤ x < a+2

Simplifying the equation, we get:

(2)^2 - a(2) + a^2 - 64 = 0

Expanding the squared term, we get:

4-2a+a^2-a2+64 = 0

Simplifying the equation, we get:

68-2a = 0

Case 3: x ≥ a+2

Simplifying the equation, we get:

(2x-2a)^2 - a(2x-2a) + a^2 - 64 = 0

Expanding the squared term, we get:

4x^2-8ax+4a2-2ax+2a^2+a2-64 = 0

Simplifying the equation, we get:

4x^2-10ax+7a2-64 = 0

Finding the Values of a

Now that we have solved the equation for each case, we can find the values of a that satisfy the equation.

Case 1: x < a-2

Simplifying the equation, we get:

4x^2-12ax+4x+a2-48 = 0

This is a quadratic equation in x, and we can solve it using the quadratic formula:

x = (-b ± √(b^2-4ac)) / 2a

In this case, a = 4, b = -12a, and c = a^2-48.

Substituting these values into the quadratic formula, we get:

x = (12a ± √((-12a)^2-4(4)(a2-48))) / 8

Simplifying the equation, we get:

x = (12a ± √(144a^2-16a2+192)) / 8

x = (12a ± √(128a^2+192)) / 8

x = (12a ± √(64(2a^2+3))) / 8

x = (12a ± 8√(2a^2+3)) / 8

x = (3a ± 2√(2a^2+3)) / 2

Case 2: a-2 ≤ x < a+2

Simplifying the equation, we get:

68-2a = 0

Solving for a, we get:

a = 34

Case 3: x ≥ a+2

Simplifying the equation, we get:

4x^2-10ax+7a2-64 = 0

This a quadratic equation in x, and we can solve it using the quadratic formula:

x = (-b ± √(b^2-4ac)) / 2a

In this case, a = 4, b = -10a, and c = 7a^2-64.

Substituting these values into the quadratic formula, we get:

x = (10a ± √((-10a)^2-4(4)(7a2-64))) / 8

Simplifying the equation, we get:

x = (10a ± √(100a^2-112a2+1024)) / 8

x = (10a ± √(-12a^2+1024)) / 8

x = (10a ± √(4(-3a^2+256))) / 8

x = (10a ± 2√(-3a^2+256)) / 8

x = (5a ± √(-3a^2+256)) / 4

Conclusion

In this article, we have solved a specific type of equation that involves absolute values, and provided a step-by-step guide on how to approach it. We have broken down the equation into three cases, and solved each case separately. We have found the values of a that satisfy the equation, and provided a clear and concise solution.

The Russian State Exam is a challenging and comprehensive exam that tests a student's knowledge and skills in various subjects, including mathematics. By practicing and mastering equations with absolute values, students can improve their chances of success in the exam.

Final Answer

The final answer is:

a = 34

Note: The final answer is only for Case 2: a-2 ≤ x < a+2. For Case 1: x < a-2 and Case 3: x ≥ a+2, the solution is a quadratic equation in x, and the values of a are not unique.

Introduction

In our previous article, we solved a specific type of equation that involves absolute values. We broke down the equation into three cases and solved each case separately. In this article, we will provide a Q&A section to help students understand the concept of absolute values and how to solve equations involving them.

Q: What is an absolute value?

A: An absolute value is the distance of a number from zero, without considering whether it's positive or negative. For example, the absolute value of 5 is 5, and the absolute value of -5 is also 5.

Q: How do I handle absolute value expressions in equations?

A: When working with absolute value expressions in equations, you need to consider two cases: when the expression is positive and when it's negative. This is because the absolute value of a number is its distance from zero, and the distance can be either positive or negative.

Q: What are the three cases for solving equations with absolute values?

A: The three cases are:

1. When the expression inside the absolute value is negative.
2. When the expression inside the absolute value is zero.
3. When the expression inside the absolute value is positive.

Q: How do I solve equations with absolute values when the expression inside the absolute value is negative?

A: When the expression inside the absolute value is negative, you need to multiply the expression by -1 to make it positive. Then, you can solve the equation as usual.

Q: How do I solve equations with absolute values when the expression inside the absolute value is zero?

A: When the expression inside the absolute value is zero, you can simply remove the absolute value sign and solve the equation as usual.

Q: How do I solve equations with absolute values when the expression inside the absolute value is positive?

A: When the expression inside the absolute value is positive, you can simply remove the absolute value sign and solve the equation as usual.

Q: What is the difference between |x| and |x-2|?

A: The expression |x| represents the distance of x from zero, while the expression |x-2| represents the distance of x from 2.

Q: How do I simplify absolute value expressions?

A: To simplify absolute value expressions, you can use the following properties:

• |x| = x when x ≥ 0
• |x| = -x when x < 0
• |x| = |y| when x = y
• |x| = |y| when x = -y

Q: Can I use absolute value expressions in inequalities?

A: Yes, you can use absolute value expressions in inequalities. However, you need to consider the same three cases as when solving equations with absolute values.

Q: How do I graph absolute value functions?

A: To graph absolute value functions, you need to consider the three cases:

1. When the expression inside the absolute value is negative.
2. When the expression inside the absolute value is zero.
3. When the expression inside the absolute value is positive.

You can use a graphing calculator or graph paper to visualize the graph.

Q: What are some common applications of absolute value equations?

A: Absolute value equations have many real-world applications, including:

• Physics: to model the of objects with constant acceleration
• Engineering: to design systems with constraints on distance or speed
• Economics: to model the behavior of economic systems with constraints on production or consumption

Conclusion

In this Q&A article, we have provided answers to common questions about solving equations with absolute values. We have covered topics such as handling absolute value expressions, solving equations with absolute values, and graphing absolute value functions. We hope that this article has helped students understand the concept of absolute values and how to apply it to solve equations and inequalities.

Final Tips

• Practice solving equations with absolute values to become more comfortable with the concept.
• Use graphing calculators or graph paper to visualize the graph of absolute value functions.
• Apply absolute value equations to real-world problems to see how they can be used to model and solve problems.

By following these tips, you can become proficient in solving equations with absolute values and apply them to a wide range of problems.