Multiplication of the sum of two elements by the difference between them

$(X + Y)\times (X - Y) = X^2 - Y^2$

This is one of the shortened multiplication formulas.

As can be seen, this formula can be used when there is a multiplication between the sum of two particular elements and the subtraction between the two elements.
Instead of presenting them as a multiplication of sum and subtraction, it can be written $X^2 - Y^2$ and it expresses exactly the same thing. In the same way, if such an expression $X^2 - Y^2$ representing the subtraction of two squared numbers is presented to you, you can write it like this: $(X + Y)\times (X - Y)$
Pay attention: the formula works both in non-algebraic expressions and in expressions that combine unknowns and numbers.

Detailed explanation

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Test yourself on difference of squares!

Solve:

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

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Let's look at an example

If we are given: $(x+4)(x-4)$
We can see that we are referring to a multiplication between the sum of two elements and the difference between them.
Therefore, we can present the same expression according to the formula in the following way:
$x^2-4^2$
$x^2-16$
In the same way, if we were given the expression:
$x^2-16$
We could express $16$ as a squared number, that is $4^2$ ,
Obtain a representation that fits the formula:
$x^2-4^2$
From here using the formula and presenting the expression in the following way:

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

If you are interested in this article, you might also be interested in the following articles:

The formula for the difference of squares

The formula for the sum of squares

The formulas that refer to two expressions to the power of 3

In the blog of Tutorela you will find a variety of articles about mathematics.

Examples and exercises with solutions for multiplying the sum of two elements by the difference between them

Exercise #1

Solve:

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

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

$4-x^2=0$

We isolate the terms and extract the root:

$4=x^2$

$x=\sqrt{4}$

$x=\pm2$

Answer

±2

Exercise #2

Complete the following exercise:

$(\sqrt{x}+\frac{1}{2})(\sqrt{x}-\frac{1}{2})=0$

Video Solution

Step-by-Step Solution

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).

Answer

$\frac{1}{4}$

Exercise #3

Solve the exercise:

^{$(x+3)(x-3)+(x+1)(x-1)=0$}

Video Solution

Step-by-Step Solution

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 $x = \pm \sqrt{5}$ .

Answer

$±\sqrt{5}$

Exercise #4

Fill in the missing element to obtain a true expression:

$(_—+3)\cdot(_—-3)=x^2-9$

Video Solution

Step-by-Step Solution

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 $x$ .

Answer

$x$

Exercise #5

Fill in the missing element to obtain a true expression:

$(x+_—)\cdot(x-_—)=x^2-121$

Video Solution

Step-by-Step Solution

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: $a = \sqrt{121} = 11$ .