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

🏆Practice difference of squares

(X+Y)×(XY)=X2Y2(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 X2Y2X^2 - Y^2 and it expresses exactly the same thing. In the same way, if such an expression X2Y2X^2 - Y^2 representing the subtraction of two squared numbers is presented to you, you can write it like this: (X+Y)×(XY)(X + Y)\times (X - Y)
Pay attention: the formula works both in non-algebraic expressions and in expressions that combine unknowns and numbers.

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Test yourself on difference of squares!

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Solve:

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

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Let's look at an example

If we are given: (x+4)(x4)(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:
x242x^2-4^2
x216x^2-16
In the same way, if we were given the expression:
x216x^2-16
We could express 1616 as a squared number, that is 424^2 ,
Obtain a representation that fits the formula:
x242x^2-4^2
From here using the formula and presenting the expression in the following way:

x242=(X4)(x+4)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)(2x)=0 (2+x)(2-x)=0

Video Solution

Step-by-Step Solution

We use the abbreviated multiplication formula:

4x2=0 4-x^2=0

We isolate the terms and extract the root:

4=x2 4=x^2

x=4 x=\sqrt{4}

x=±2 x=\pm2

Answer

±2

Exercise #2

Complete the following exercise:

(x+12)(x12)=0 (\sqrt{x}+\frac{1}{2})(\sqrt{x}-\frac{1}{2})=0

Video Solution

Step-by-Step Solution

To solve the equation (x+12)(x12)=0(\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: x+12=0\sqrt{x} + \frac{1}{2} = 0
    Solving for xx, subtract 12\frac{1}{2} from both sides:
    x=12\sqrt{x} = -\frac{1}{2}
    Squaring both sides, we get:
    x=(12)2=14x = \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: x12=0\sqrt{x} - \frac{1}{2} = 0
    Solving for xx, add 12\frac{1}{2} to both sides:
    x=12\sqrt{x} = \frac{1}{2}
    Squaring both sides, we have:
    x=(12)2=14x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.

Therefore, the solution to the equation (x+12)(x12)=0(\sqrt{x} + \frac{1}{2})(\sqrt{x} - \frac{1}{2}) = 0 is x=14x = \frac{1}{4}.

Upon reviewing the provided choices, the correct answer that matches our solution is: 14 \frac{1}{4} (Option 2).

Answer

14 \frac{1}{4}

Exercise #3

Solve the exercise:

(x+3)(x3)+(x+1)(x1)=0 (x+3)(x-3)+(x+1)(x-1)=0

Video Solution

Step-by-Step Solution

To solve the equation (x+3)(x3)+(x+1)(x1)=0 (x+3)(x-3) + (x+1)(x-1) = 0 , we will employ the difference of squares formula.

Step 1: Simplify (x+3)(x3)(x+3)(x-3) using the difference of squares:
(x+3)(x3)=x232=x29(x+3)(x-3) = x^2 - 3^2 = x^2 - 9.

Step 2: Simplify (x+1)(x1)(x+1)(x-1) using the difference of squares:
(x+1)(x1)=x212=x21(x+1)(x-1) = x^2 - 1^2 = x^2 - 1.

Step 3: Substitute the simplified expressions back into the original equation:
x29+x21=0x^2 - 9 + x^2 - 1 = 0.

Step 4: Combine like terms:
2x210=02x^2 - 10 = 0.

Step 5: Simplify the equation by factoring or isolating x2x^2:
Divide through by 2 to get x25=0x^2 - 5 = 0.

Step 6: Solve for x2x^2:
x2=5x^2 = 5.

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

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

Answer

±5 ±\sqrt{5}

Exercise #4

Fill in the missing element to obtain a true expression:

(x+)(x)=x2121 (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)(xa)=x2a2 (x+a)(x-a) = x^2 - a^2 .
  • Determine the values of a a to fill in the blanks.

Now, let's work through each step:
Step 1: The expression given is (x+_)(x_)=x2121 (x+\_—)\cdot(x-\_—) = x^2-121 . Recognize that x2121 x^2 - 121 is a difference of squares.
Step 2: We know from the difference of squares formula that a2=121 a^2 = 121 .
Step 3: Solve for a a by taking the square root of both sides: a=121=11 a = \sqrt{121} = 11 .

This means the expression becomes: (x+11)(x11)=x2121 (x+11)(x-11) = x^2 - 121 .

Therefore, the missing element is 11 11 .

Answer

11

Exercise #5

Fill in the missing element to obtain a true expression:

(+3)(3)=x29 (_—+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)(ab)=a2b2 (a + b)(a - b) = a^2 - b^2 . Given the equation (+3)(3)=x29(_ + 3)(_- 3) = x^2 - 9, we can compare it to the formula:

  • a2=x2 a^2 = x^2 implies a=x a = x .
  • b2=9 b^2 = 9 implies b=3 b = 3 .

This means the expression (+3)(3)(_ + 3)(_- 3) should represent (x+3)(x3)(x + 3)(x - 3), satisfying the equation through the difference of squares formula.

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

Answer

x x

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