What Is The Solution To $-4|-2x + 6| = -24$?A. $x = 0$ B. $x = 0$ Or $x = -6$ C. $x = 0$ Or $x = 6$ D. No Solution

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What is the Solution to βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24?

Understanding the Problem

The given equation is βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24. To solve this equation, we need to isolate the variable xx. The equation involves absolute value, which means we need to consider both positive and negative cases.

Breaking Down the Absolute Value

The absolute value of a quantity is its distance from zero on the number line. In this case, the absolute value is βˆ£βˆ’2x+6∣|-2x + 6|. To remove the absolute value, we need to consider two cases:

  • Case 1: βˆ’2x+6β‰₯0-2x + 6 \geq 0
  • Case 2: βˆ’2x+6<0-2x + 6 < 0

Case 1: βˆ’2x+6β‰₯0-2x + 6 \geq 0

In this case, the absolute value can be rewritten as βˆ£βˆ’2x+6∣=βˆ’2x+6|-2x + 6| = -2x + 6. Substituting this into the original equation, we get:

βˆ’4(βˆ’2x+6)=βˆ’24-4(-2x + 6) = -24

Expanding the left-hand side, we get:

8xβˆ’24=βˆ’248x - 24 = -24

Adding 24 to both sides, we get:

8x=08x = 0

Dividing both sides by 8, we get:

x=0x = 0

Case 2: βˆ’2x+6<0-2x + 6 < 0

In this case, the absolute value can be rewritten as βˆ£βˆ’2x+6∣=2xβˆ’6|-2x + 6| = 2x - 6. Substituting this into the original equation, we get:

βˆ’4(2xβˆ’6)=βˆ’24-4(2x - 6) = -24

Expanding the left-hand side, we get:

βˆ’8x+24=βˆ’24-8x + 24 = -24

Subtracting 24 from both sides, we get:

βˆ’8x=βˆ’48-8x = -48

Dividing both sides by -8, we get:

x=6x = 6

Combining the Cases

We have found two possible solutions: x=0x = 0 and x=6x = 6. However, we need to check if these solutions satisfy the condition βˆ’2x+6<0-2x + 6 < 0.

For x=0x = 0, we have βˆ’2(0)+6=6-2(0) + 6 = 6, which is not less than 0. Therefore, x=0x = 0 is not a valid solution.

For x=6x = 6, we have βˆ’2(6)+6=βˆ’6-2(6) + 6 = -6, which is less than 0. Therefore, x=6x = 6 is a valid solution.

Conclusion

The solution to the equation βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24 is x=6x = 6. This is the only valid solution that satisfies the condition βˆ’2x+6<0-2x + 6 < 0.

Answer

The correct answer is C. x=0x = 0 or x=6x = 6 is incorrect, the correct answer is only x=6x = 6
Q&A: Understanding the Solution to βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24

Q: What is the main concept behind solving the equation βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24?

A: The main concept behind solving this equation is to understand the properties of absolute value and how to remove it from the equation. Absolute value represents the distance of a quantity from zero on the number line, and it can be positive or negative.

Q: How do you handle absolute value in an equation?

A: To handle absolute value in an equation, you need to consider two cases:

  • Case 1: The quantity inside the absolute value is non-negative.
  • Case 2: The quantity inside the absolute value is negative.

Q: What is the difference between the two cases?

A: In Case 1, the absolute value can be rewritten as the quantity inside the absolute value. In Case 2, the absolute value can be rewritten as the negative of the quantity inside the absolute value.

Q: How do you determine which case to use?

A: To determine which case to use, you need to check the sign of the quantity inside the absolute value. If it is non-negative, use Case 1. If it is negative, use Case 2.

Q: What is the solution to the equation βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24?

A: The solution to the equation βˆ’4βˆ£βˆ’2x+6∣=βˆ’24-4|-2x + 6| = -24 is x=6x = 6. This is the only valid solution that satisfies the condition βˆ’2x+6<0-2x + 6 < 0.

Q: Why is x=0x = 0 not a valid solution?

A: x=0x = 0 is not a valid solution because it does not satisfy the condition βˆ’2x+6<0-2x + 6 < 0. When x=0x = 0, βˆ’2x+6=6-2x + 6 = 6, which is not less than 0.

Q: What is the importance of checking the condition βˆ’2x+6<0-2x + 6 < 0?

A: Checking the condition βˆ’2x+6<0-2x + 6 < 0 is important because it ensures that the solution satisfies the original equation. If the solution does not satisfy the condition, it is not a valid solution.

Q: Can you provide an example of a similar equation that can be solved using the same method?

A: Yes, an example of a similar equation is βˆ’3∣xβˆ’4∣=9-3|x - 4| = 9. To solve this equation, you can follow the same steps as before: consider two cases, rewrite the absolute value, and solve for the variable.

Q: What are some common mistakes to avoid when solving equations with absolute value?

A: Some common mistakes to avoid when solving equations with absolute value include:

  • Failing to consider both cases
  • Not rewriting the absolute value correctly
  • Not checking the condition that the quantity inside the absolute value must satisfy

Q: How can you apply the concept of absolute value to real-world problems?

A: The concept of absolute value can be applied to real-world problems in various fields, such as physics, engineering, and finance. For example, in physics, absolute value can be used to represent the distance between two objects. In finance, absolute value can be used to represent the amount of money that is owed or received.