Absolute Value Math Problems: Solve And Learn!

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Hey guys! Let's dive into some absolute value problems today. We'll break down each step, so you can understand the process and ace similar questions in the future. Math can be fun, especially when you get the hang of it!

Problem 1: Calculating with Absolute Values and Variables

Our first problem involves a bit of algebra and absolute values. The question is:

Calculate (βˆ’4)imesβˆ£βˆ’a∣2βˆ’βˆ£βˆ’a∣imes∣a∣(-4) imes |{-a}|^{2} - |{-a}| imes |{a}|.

This looks a little intimidating at first, but don't worry, we'll take it step by step. The key here is understanding what absolute value means and how to handle variables within absolute value signs.

Understanding Absolute Value

First, let's recap what absolute value is. The absolute value of a number is its distance from zero on the number line. It's always non-negative. So, ∣5∣=5|5| = 5 and βˆ£βˆ’5∣=5|-5| = 5. The absolute value essentially strips away the negative sign if there is one.

Breaking Down the Expression

Now, let's look at our expression: (βˆ’4)imesβˆ£βˆ’a∣2βˆ’βˆ£βˆ’a∣imes∣a∣(-4) imes |{-a}|^{2} - |{-a}| imes |{a}|.

We have absolute values of βˆ’a-a and aa. Remember, the absolute value of βˆ’a-a is the same as the absolute value of aa, because they are the same distance from zero. We can write this as βˆ£βˆ’a∣=∣a∣|-a| = |a|. Let's represent ∣a∣|a| with a simpler variable, say xx. So, x=∣a∣=βˆ£βˆ’a∣x = |a| = |-a|.

Now our expression looks like this: (βˆ’4)imesx2βˆ’ximesx(-4) imes x^{2} - x imes x.

Simplifying the Expression

Let's simplify further. We have:

(βˆ’4)imesx2βˆ’x2(-4) imes x^{2} - x^{2}

This simplifies to:

βˆ’4x2βˆ’x2-4x^{2} - x^{2}

Combining the terms, we get:

βˆ’5x2-5x^{2}

Substituting Back

Remember that x=∣a∣x = |a|, so we substitute that back in:

βˆ’5∣a∣2-5|a|^{2}

So, the final simplified expression is βˆ’5∣a∣2-5|a|^{2}.

Key Steps and Takeaways

  • Understand Absolute Value: The absolute value of a number is its distance from zero.
  • Simplify Variables: Using a temporary variable like 'x' can make the expression easier to handle.
  • Substitute Back: Don't forget to substitute the original variable back into the final expression.

This problem highlights how simplifying expressions and understanding the properties of absolute values can lead to a clear solution. Now, let's move on to the second problem!

Problem 2: Evaluating a Function with Absolute Values

Okay, so the next question involves a function with absolute values. These can seem tricky, but once you understand how to plug in values, they become much simpler. Here’s the question:

Given the absolute value function f(x)=∣2xβˆ’3∣f(x) = |2x - 3|, find the value of f(0) + f( rac{1}{2}) - f(-4) imes f(2).

Understanding Function Notation

First off, let’s quickly recap function notation. When we write f(x)f(x), it means we have a function named 'f' that takes 'x' as an input. To find f(0)f(0), we simply replace every 'x' in the function's expression with 0. Similarly, for f( rac{1}{2}), f(βˆ’4)f(-4), and f(2)f(2), we replace 'x' with rac{1}{2}, -4, and 2, respectively.

Evaluating the Function at Different Points

Now, let's calculate each part:

  1. Finding f(0):

    • f(0)=∣2(0)βˆ’3∣f(0) = |2(0) - 3|
    • f(0)=∣0βˆ’3∣f(0) = |0 - 3|
    • f(0)=βˆ£βˆ’3∣f(0) = |-3|
    • f(0)=3f(0) = 3

  2. Finding f(1/2):

    • f( rac{1}{2}) = |2( rac{1}{2}) - 3|
    • f( rac{1}{2}) = |1 - 3|
    • f( rac{1}{2}) = |-2|
    • f( rac{1}{2}) = 2

  3. Finding f(-4):

    • f(βˆ’4)=∣2(βˆ’4)βˆ’3∣f(-4) = |2(-4) - 3|
    • f(βˆ’4)=βˆ£βˆ’8βˆ’3∣f(-4) = |-8 - 3|
    • f(βˆ’4)=βˆ£βˆ’11∣f(-4) = |-11|
    • f(βˆ’4)=11f(-4) = 11

  4. Finding f(2):

    • f(2)=∣2(2)βˆ’3∣f(2) = |2(2) - 3|
    • f(2)=∣4βˆ’3∣f(2) = |4 - 3|
    • f(2)=∣1∣f(2) = |1|
    • f(2)=1f(2) = 1

Putting It All Together

Now that we have the values of f(0)f(0), f( rac{1}{2}), f(βˆ’4)f(-4), and f(2)f(2), we can plug them into the original expression:

f(0) + f( rac{1}{2}) - f(-4) imes f(2)

Substitute the values:

3+2βˆ’11imes13 + 2 - 11 imes 1

Calculating the Final Answer

Now, let's do the arithmetic:

3+2βˆ’11=5βˆ’11=βˆ’63 + 2 - 11 = 5 - 11 = -6

So, the final answer is -6.

Key Steps and Takeaways

  • Function Notation: Understand how to substitute values into a function.
  • Evaluate Step-by-Step: Calculate each function value separately to avoid confusion.
  • Order of Operations: Remember to follow the correct order of operations (PEMDAS/BODMAS) when combining the values.

This problem illustrates how to evaluate functions with absolute values by breaking them down into smaller, manageable steps. Let's wrap things up with some final thoughts.

Final Thoughts and Tips for Absolute Value Problems

Guys, we've tackled two pretty interesting absolute value problems today. Remember, the key to mastering these types of questions is understanding the basics and practicing consistently. Here are a few extra tips to keep in mind:

  • Always consider both positive and negative cases: When dealing with absolute values, remember that the expression inside the absolute value can be either positive or negative. This is especially important when solving equations or inequalities involving absolute values.
  • Use a number line: Visualizing absolute values on a number line can help you understand the concept better and solve problems more intuitively.
  • Practice, practice, practice: The more problems you solve, the more comfortable you'll become with absolute values.

Math can sometimes feel like a challenge, but with a bit of patience and the right approach, you can totally nail it. Keep up the great work, and I'll catch you in the next math adventure!

In Summary

  1. For the first problem, (βˆ’4)imesβˆ£βˆ’a∣2βˆ’βˆ£βˆ’a∣imes∣a∣(-4) imes |{-a}|^{2} - |{-a}| imes |{a}|, we simplified the expression to βˆ’5∣a∣2-5|a|^{2}.
  2. For the second problem, given f(x)=∣2xβˆ’3∣f(x) = |2x - 3|, we found that f(0) + f( rac{1}{2}) - f(-4) imes f(2) = -6.

Remember to review these steps and try similar problems to solidify your understanding. You've got this!