Determine the Validity: Is x^2 - 81 Over (x-9)(x+9) Equal to 1?

Rational Expressions with Domain Restrictions

Does the following equation have a true or false value?

x281(x9)(x+9)=1 \frac{x^2-81}{(x-9)(x+9)}=1

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Is the equation correct?
00:05 Let's break down 81 into 9 squared
00:14 Let's use the shortened multiplication formulas
00:23 Let's reduce what we can
00:28 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Does the following equation have a true or false value?

x281(x9)(x+9)=1 \frac{x^2-81}{(x-9)(x+9)}=1

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Recognize and factor the expression in the numerator.
  • Step 2: Simplify the fraction by canceling common factors.
  • Step 3: Determine restrictions on the variable x x .
  • Step 4: Analyze if and when the equation holds true.

Now, let's work through each step:

Step 1: The numerator x281 x^2 - 81 can be factored as a difference of squares: (x9)(x+9) (x-9)(x+9) .

Step 2: Substitute this factorization into the equation:
(x9)(x+9)(x9)(x+9)=1\frac{(x-9)(x+9)}{(x-9)(x+9)} = 1.

Step 3: Simplify the fraction by canceling the common terms, giving 1=1 1 = 1 , which is always true, except where the expression is undefined.

Step 4: The expression is undefined when the denominator is zero, i.e., when x9=0 x - 9 = 0 or x+9=0 x + 9 = 0 . Thus, x9 x \neq 9 and x9 x \neq -9 .

In conclusion, the given equation x281(x9)(x+9)=1 \frac{x^2-81}{(x-9)(x+9)}=1 is True only when x±9 x \ne \pm9 .

3

Final Answer

True only when x±9 x\ne\pm9 .

Key Points to Remember

Essential concepts to master this topic
  • Factoring: Recognize x281 x^2 - 81 as difference of squares pattern
  • Technique: Cancel common factors: (x9)(x+9)(x9)(x+9)=1 \frac{(x-9)(x+9)}{(x-9)(x+9)} = 1
  • Check: Verify domain restrictions where denominator equals zero: x ≠ ±9 ✓

Common Mistakes

Avoid these frequent errors
  • Ignoring domain restrictions after simplification
    Don't just simplify to 1 = 1 and say it's always true = wrong conclusion! The original expression is undefined when x = ±9, creating holes in the domain. Always identify where the denominator equals zero before simplifying.

Practice Quiz

Test your knowledge with interactive questions

Solve:

\( (2+x)(2-x)=0 \)

FAQ

Everything you need to know about this question

Why can't I just cancel everything and say the equation is always true?

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Great question! While the simplified form 1 = 1 is always true, you must remember the original expression had restrictions. The fraction x281(x9)(x+9) \frac{x^2-81}{(x-9)(x+9)} doesn't exist when x = ±9!

What happens at x = 9 or x = -9?

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At these values, both the numerator and denominator equal zero, creating the indeterminate form 00 \frac{0}{0} . This means the expression is undefined at these points, even though it simplifies to 1 everywhere else.

How do I identify domain restrictions quickly?

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Look at the original denominator before any simplification. Set each factor equal to zero and solve:

  • x - 9 = 0 → x = 9
  • x + 9 = 0 → x = -9

These are your restricted values!

Is this different from removable discontinuities in graphing?

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Exactly! This creates holes in the graph at x = ±9. The function equals 1 everywhere except these two points, where it's undefined. Think of it as a horizontal line y = 1 with two holes punched out.

What if the question asked if the equation is true or false?

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This is a trick question! It's neither simply true nor false. The correct answer is "True only when x ≠ ±9" because you must account for the domain restrictions of the original expression.

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