Difference of squares - Examples, Exercises and Solutions

We use the abbreviated multiplication formula:

$4-x^2=0$

We isolate the terms and extract the root:

$4=x^2$

$x=\sqrt{4}$

$x=\pm2$

To solve the equation $(\sqrt{x} + \frac{1}{2})(\sqrt{x} - \frac{1}{2}) = 0$, we can apply the zero-product property, which tells us that if a product of two factors is zero, at least one of the factors must be zero.

Let us proceed with each factor:

• First Factor: $\sqrt{x} + \frac{1}{2} = 0$
  Solving for $x$, subtract $\frac{1}{2}$ from both sides:
  $\sqrt{x} = -\frac{1}{2}$
  Squaring both sides, we get:
  $x = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$.
  However, since the square root should be zero or positive, this case does not yield a real solution.
• Second Factor: $\sqrt{x} - \frac{1}{2} = 0$
  Solving for $x$, add $\frac{1}{2}$ to both sides:
  $\sqrt{x} = \frac{1}{2}$
  Squaring both sides, we have:
  $x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

Therefore, the solution to the equation $(\sqrt{x} + \frac{1}{2})(\sqrt{x} - \frac{1}{2}) = 0$ is $x = \frac{1}{4}$.

Upon reviewing the provided choices, the correct answer that matches our solution is: $\frac{1}{4}$ (Option 2).

To solve the equation $(x+3)(x-3) + (x+1)(x-1) = 0$, we will employ the difference of squares formula.

Step 1: Simplify $(x+3)(x-3)$ using the difference of squares:
$(x+3)(x-3) = x^2 - 3^2 = x^2 - 9$.

Step 2: Simplify $(x+1)(x-1)$ using the difference of squares:
$(x+1)(x-1) = x^2 - 1^2 = x^2 - 1$.

Step 3: Substitute the simplified expressions back into the original equation:
$x^2 - 9 + x^2 - 1 = 0$.

Step 4: Combine like terms:
$2x^2 - 10 = 0$.

Step 5: Simplify the equation by factoring or isolating $x^2$:
Divide through by 2 to get $x^2 - 5 = 0$.

Step 6: Solve for $x^2$:
$x^2 = 5$.

Step 7: Solve for $x$ by taking the square root of both sides:
$x = \pm \sqrt{5}$.

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

To solve this problem, let's use the difference of squares formula, which is $(a + b)(a - b) = a^2 - b^2$. Given the equation $(_ + 3)(_- 3) = x^2 - 9$, we can compare it to the formula:

• $a^2 = x^2$ implies $a = x$.
• $b^2 = 9$ implies $b = 3$.

This means the expression $(_ + 3)(_- 3)$ should represent $(x + 3)(x - 3)$, satisfying the equation through the difference of squares formula.

Thus, the missing element to obtain a correct expression is $x$.

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

• Identify the given expression as a difference of squares.
• Apply the formula for finding the missing term in $(x+a)(x-a) = x^2 - a^2$.
• Determine the values of $a$ to fill in the blanks.

Now, let's work through each step:
Step 1: The expression given is $(x+\_—)\cdot(x-\_—) = x^2-121$. Recognize that $x^2 - 121$ is a difference of squares.
Step 2: We know from the difference of squares formula that $a^2 = 121$.
Step 3: Solve for $a$ by taking the square root of both sides: $a = \sqrt{121} = 11$.

This means the expression becomes: $(x+11)(x-11) = x^2 - 121$.

Therefore, the missing element is $11$.