Question

Alice made the conjecture (a+b)² = a² + b². Which values of a and b are NOT counterexamples in the conjecture?

F. a = 0, b = 1
G. a = 1, b = 1
H. a = -1, b = 1
J. a = -1, b = 2

Answer 1

The values a = 0, b = 1 (option F) are NOT counterexamples to the conjecture (a+b)² = a² + b², because when a is zero, the conjecture holds true.

Step-by-step explanation:

Alice made the conjecture (a+b)² = a² + b². To determine which values of a and b are NOT counterexamples to the conjecture, we must verify when the equation holds true. Using the binomial expansion, we know that (a+b)² = a² + 2ab + b². Therefore, the only way for Alice's conjecture to be true is if 2ab equals zero, which occurs if at least one of the variables, a or b, is zero.

Among the given options:

• F. a = 0, b = 1
  
• G. a = 1, b = 1
  
• H. a = -1, b = 1
  
• J. a = -1, b = 2

Option F does not provide a counterexample because when a is zero, the conjecture holds true (0² + 1² = 0 + 1 = 1, which is equal to (0+1)²).