Difference of squares: Using fractions

To solve the problem, begin with simplifying the left-hand side of the equation:

$(x+7)(x-7) = x^2 - 49$.

Thus, the original equation $\frac{(x+7)(x-7)}{3} = -11 - x^2$ simplifies to:

$\frac{x^2 - 49}{3} = -11 - x^2$.

Multiplying every term by 3 to clear the fraction, we obtain:

$x^2 - 49 = -33 - 3x^2$.

Add $3x^2$ to both sides to consolidate $x^2$ terms on one side:

$x^2 + 3x^2 = -33 + 49$.

This simplifies to:

$4x^2 = 16$.

Divide by 4 on both sides:

$x^2 = 4$.

Taking the square root of both sides provides:

$x = \pm 2$.

Therefore, the solution to the problem is $x = \pm 2$, corresponding to the choice labeled

±2

To solve the equation $\frac{2x^2-32}{8} = \frac{x+4}{2}$, let's proceed through these steps:

• Step 1: Simplify and eliminate fractions by cross-multiplying.
• Step 2: Rearrange and simplify the resulting equation.
• Step 3: Solve for the variable $x$.

Now, let's work through each step:

Step 1: Cross-multiply to eliminate the fractions.
We perform cross multiplication as follows:

This gives us:

Step 2: Rearrange and simplify the equation.
Move all terms to one side to set the equation to zero:

Simplify by dividing the entire equation by 4:

Step 3: Solve the quadratic equation by factoring.
Factor the quadratic equation:

Set each factor to zero and solve for $x$:

• $x - 6 = 0 \Rightarrow x = 6$
• $x + 4 = 0 \Rightarrow x = -4$

Considering the given multiple-choice answers, the correct solution is:

$x = 6$

Therefore, the solution to the problem is $x = 6$.

First we rearrange the equation and set it to 0

$4x^2-3-6=0$

$4x^2-9=0$

We then apply the shortcut multiplication formula:

$4(x^2-\frac{9}{4})=0$

$x^2-(\frac{3}{2})^2=0$

$(x-\frac{3}{2})(x+\frac{3}{2})=0$

$x=\pm\frac{3}{2}$

To solve this problem, we need to find the values of $x$ that make the equation $\frac{x(x^2-9)}{x^2+3x}=0$ true. The steps are as follows:

• Step 1: Simplify the Numerator
  The numerator is $x(x^2 - 9)$. Recognize $x^2 - 9$ as a difference of squares, which can be factored to $(x-3)(x+3)$. Thus, the numerator becomes $x(x-3)(x+3)$.

• Step 2: Simplify the Denominator
  The denominator is $x^2 + 3x$, which can be factored as $x(x + 3)$.

• Step 3: Rewrite the Equation
  Now, the equation is rewritten as:
  $\frac{x(x-3)(x+3)}{x(x+3)} = 0$

• Step 4: Cancel Common Factors
  Assuming $x \neq 0$ (since division by zero is undefined), cancel $x$ and $(x+3)$:
  $\frac{x-3}{1} = 0$

• Step 5: Solve for $x$
  The reduced equation $x - 3 = 0$ gives the solution $x = 3$.

• Step 6: Check for Restrictions
  We previously canceled $x$ and $x+3$, so $x \neq 0$ and $x \neq -3$ must be considered as part of the domain.

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

This corresponds to choice 2 in the given multiple-choice options.

To solve this problem, follow these steps:

• Step 1: Factorize the terms involving squares. Notice that $x^2 - 64 = (x-8)(x+8)$ and $64 - x^2 = (8-x)(8+x)$.
• Step 2: Rewrite both sides of the equation using these factorizations:

$\frac{(x-8)(x+8)}{x-8} = \frac{17(x+8)}{(8-x)(8+x)}$

Upon simplifying, the left side becomes $x+8$ because the $(x-8)$ term cancels out, as long as $x \neq 8$.

$x + 8 = \frac{17(x+8)}{-(x-8)(x+8)}$

• Step 3: Cancel $(x+8)$ from both numerator and denominator on the right side (assuming $x \neq -8$).
• Step 4: This simplifies to:

$x + 8 = \frac{-17}{x-8}$

• Step 5: Multiply both sides by $(x-8)$ to eliminate the denominator:

$(x+8)(x-8) = -17$

$x^2 - 64 = -17$

• Step 6: Rearrange and solve for $x$:

$x^2 = 47$

$x = \pm \sqrt{47}$

Therefore, the solution to the problem is $x = \pm \sqrt{47}$.