Solve (a+b)(a-b) = (a+b)²: Finding Values of a and b

Algebraic Identities with Factoring Solutions

Find a,b a ,b such that:

(a+b)(ab)=(a+b)2 (a+b)(a-b)=(a+b)^2

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find A,B such that the equation is satisfied
00:03 Use the shortened multiplication formulas to expand the parentheses
00:17 Simplify what we can
00:33 Arrange the equation so that 0 is on one side
00:43 Collect terms
00:53 Factor the expression
00:57 Find the common factor and take it out of the parentheses
01:10 Now find which solutions make the equation equal to zero
01:14 Set each factor in the product to 0 and solve
01:17 This is one solution
01:21 Set the second factor of the product to 0
01:26 And this is the second solution
01:29 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find a,b a ,b such that:

(a+b)(ab)=(a+b)2 (a+b)(a-b)=(a+b)^2

2

Step-by-step solution

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

  • Step 1: Expand both sides of the given equation.
  • Step 2: Compare and simplify the resulting expressions.
  • Step 3: Solve for aa and bb.

Now, let's work through each step:
Step 1: The given equation is (a+b)(ab)=(a+b)2(a+b)(a-b) = (a+b)^2. Let's expand both sides:
- Left side: (a+b)(ab)=a2b2(a+b)(a-b) = a^2 - b^2 based on the difference of squares formula.
- Right side: (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 using the square of a sum formula.

Step 2: Setting the expanded forms equal gives us:
a2b2=a2+2ab+b2a^2 - b^2 = a^2 + 2ab + b^2.

Step 3: Simplify and solve the equation:
- Subtract a2a^2 from both sides: b2=2ab+b2-b^2 = 2ab + b^2.
- Add b2b^2 to both sides: 0=2ab+2b20 = 2ab + 2b^2.
- Factor the right-hand side: 0=2b(a+b)0 = 2b(a + b).

This gives us two possible conditions:
1) 2b=02b = 0, which implies b=0b = 0.
2) a+b=0a + b = 0, which implies a=ba = -b.

Since a=ba = -b satisfies the equation for any aa if bb is not zero, and when b=0b = 0, the equation simplifies to 0=00 = 0, both conditions are valid.

Therefore, the solutions are a=ba = -b or b=0b = 0.

In conclusion, the answer is: a=b a=-b or 0=b 0=b .

3

Final Answer

a=b a=-b or

0=b 0=b

Key Points to Remember

Essential concepts to master this topic
  • Identity Recognition: Use difference of squares and perfect square formulas
  • Technique: Factor out common terms: 2b(a + b) = 0
  • Check: Substitute solutions back into original equation to verify ✓

Common Mistakes

Avoid these frequent errors
  • Incorrectly expanding algebraic expressions
    Don't expand (a+b)(ab)(a+b)(a-b) as a2+b2a^2 + b^2 = wrong identity! This misses the negative sign and gives incorrect results. Always remember the difference of squares: (a+b)(ab)=a2b2(a+b)(a-b) = a^2 - b^2.

Practice Quiz

Test your knowledge with interactive questions

Solve:

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

FAQ

Everything you need to know about this question

Why does the equation have two different solutions?

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When we factor 2b(a+b)=02b(a + b) = 0, we get a product equal to zero. This means either factor can be zero: 2b = 0 (so b = 0) or a + b = 0 (so a = -b).

Can both a and b be zero at the same time?

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Yes! If a = 0 and b = 0, then both conditions are satisfied. This makes a=ba = -b true (0 = -0) and also b=0b = 0 true.

How do I know which algebraic identity to use?

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Look at the structure: (a+b)(ab)(a+b)(a-b) is the difference of squares pattern, while (a+b)2(a+b)^2 is the perfect square pattern. Memorize these key formulas!

What if I expand everything first instead of factoring?

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That works too! You'll get a2b2=a2+2ab+b2a^2 - b^2 = a^2 + 2ab + b^2, which simplifies to the same result. Both methods are valid - choose what feels easier for you.

Are there other values of a and b that work?

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No! The only solutions are when b=0b = 0 (with any value of a) or when a=ba = -b (with any non-zero b). These cover all possible solutions.

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