Difference of squares: Equations with denominators

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

• Step 1: Factor the numerator using the difference of squares formula.
• Step 2: Set the factored numerator equal to zero to find potential solutions for $x$.
• Step 3: Ensure none of these solutions result in the denominator being zero.

Now, let's work through each step:
Step 1: The numerator $x^2 - 9$ can be factored as $(x - 3)(x + 3)$.
Step 2: Set the factored numerator to zero: $(x - 3)(x + 3) = 0$.
This gives two potential solutions: $x - 3 = 0$ or $x + 3 = 0$. Solving these equations, we get $x = 3$ or $x = -3$.
Step 3: Verify that these solutions do not result in division by zero:
- For $x = 3$, the denominator $x - 3 = 0$, which means division by zero occurs, so it is not a valid solution.
- For $x = -3$, the denominator $x - 3$ is not zero, as $(-3 - 3) = -6$, hence $x = -3$ is a valid solution.

Therefore, the solution to the problem is $x = -3$.

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 $x$.
• Step 4: Analyze if and when the equation holds true.

Now, let's work through each step:

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

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

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

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

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

To solve the equation $\frac{x^2 + 4}{3} = 3(x-2)(x+2)$, we will follow these steps:

• Step 1: Remove the fraction by multiplying both sides by 3.
• Step 2: Expand and simplify the right side using the difference of squares.
• Step 3: Rearrange into a standard quadratic form.
• Step 4: Solve the quadratic equation.

Let's begin:

Step 1: Multiply both sides by 3 to eliminate the fraction:

$x^2 + 4 = 9(x-2)(x+2)$

Step 2: Recognize that $(x-2)(x+2)$ is a difference of squares:

$(x-2)(x+2) = x^2 - 4$

Then we have:

$x^2 + 4 = 9(x^2 - 4)$

Step 3: Distribute the 9 on the right side:

$x^2 + 4 = 9x^2 - 36$

Step 4: Rearrange this into a standard quadratic form:

$x^2 - 9x^2 = -36 - 4$

Simplify:

$-8x^2 = -40$

Divide everything by -8 to solve for $x^2$:

$x^2 = 5$

Step 5: Solve by taking the square root of both sides:

$x = \pm \sqrt{5}$

Therefore, the solution to the problem is $x = \pm \sqrt{5}$.

To solve this equation, we follow these steps:

• Step 1: Identify the original equation as $\frac{-7}{x+4} = x-4$.
• Step 2: Eliminate the fraction by multiplying both sides by $x+4$:
  $-7 = (x-4)(x+4)$.
• Step 3: Recognize the right-hand side as a difference of squares:
  $-7 = x^2 - 16$.
• Step 4: Rearrange the equation into a standard quadratic form:
  $x^2 - 16 + 7 = 0$ which simplifies to $x^2 - 9 = 0$.
• Step 5: Factor the quadratic expression:
  $(x-3)(x+3) = 0$.
• Step 6: Use the zero product property to find potential solutions:
  $x - 3 = 0$ which gives $x = 3$, and
  $x + 3 = 0$ which gives $x = -3$.
• Step 7: Validate that neither solution makes the denominator zero, as $x \neq -4$.

Thus, the solutions to the equation $\frac{-7}{x+4} = x-4$ are $\pm3$.