Difference of Squares Practice Problems with Solutions

Master the difference of squares formula (x+y)(x-y)=x²-y² with step-by-step practice problems. Learn to factor and expand expressions using this key algebraic identity.

📚Master the Difference of Squares Formula Through Practice
  • Apply the formula (X+Y)(X-Y)=X²-Y² to expand algebraic expressions
  • Factor expressions like x²-16 into (x+4)(x-4) form
  • Recognize when expressions fit the difference of squares pattern
  • Solve problems combining numbers and algebraic variables
  • Use shortened multiplication formulas to simplify calculations
  • Master both expansion and factoring directions of the formula

Understanding Difference of squares

Complete explanation with examples

(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.

Detailed explanation

Practice Difference of squares

Test your knowledge with 12 quizzes

Fill in the missing element to obtain a true expression:

\( 2x^2-_{_—}=2(x-4)\cdot(x+4) \)

Examples with solutions for Difference of squares

Step-by-step solutions included
Exercise #1

Solve:

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

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

Video Solution
Exercise #2

Complete the following exercise:

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

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}

Video Solution
Exercise #3

Solve the exercise:

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

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}

Video Solution
Exercise #4

Fill in the missing element to obtain a true expression:

(+3)(3)=x29 (_—+3)\cdot(_—-3)=x^2-9

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

Video Solution
Exercise #5

Fill in the missing element to obtain a true expression:

(x+)(x)=x2121 (x+_—)\cdot(x-_—)=x^2-121

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

Video Solution

Frequently Asked Questions

What is the difference of squares formula?

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The difference of squares formula is (X+Y)(X-Y)=X²-Y². This shortened multiplication formula allows you to quickly expand the product of a sum and difference, or factor the difference of two perfect squares.

How do you factor x²-25 using difference of squares?

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To factor x²-25: 1) Recognize it as x²-5² (difference of squares pattern), 2) Apply the formula X²-Y²=(X+Y)(X-Y), 3) Get x²-5²=(x+5)(x-5).

When can you use the difference of squares formula?

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You can use the difference of squares formula when you have two perfect squares separated by subtraction (like a²-b²), or when multiplying expressions in the form (a+b)(a-b). It works with both numbers and algebraic variables.

What are common mistakes with difference of squares?

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Common mistakes include: • Trying to use the formula with addition (a²+b² doesn't factor), • Forgetting to check if terms are perfect squares, • Sign errors when expanding or factoring, • Not recognizing disguised forms like 4x²-9.

How do you expand (2x+3)(2x-3)?

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Using the difference of squares formula: (2x+3)(2x-3) = (2x)²-3² = 4x²-9. The middle terms cancel out, leaving only the difference of the squared terms.

Can difference of squares work with fractions and decimals?

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Yes, the difference of squares formula works with any terms that can be squared. For example: (1/2+x)(1/2-x) = (1/2)²-x² = 1/4-x², and (0.5+y)(0.5-y) = 0.25-y².

What's the difference between sum of squares and difference of squares?

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The difference of squares (a²-b²) can be factored as (a+b)(a-b), while the sum of squares (a²+b²) cannot be factored using real numbers. Only the difference of squares has a factoring formula.

How do you solve equations using difference of squares?

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To solve equations like x²-16=0: 1) Factor using difference of squares: (x+4)(x-4)=0, 2) Set each factor to zero: x+4=0 or x-4=0, 3) Solve: x=-4 or x=4.

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