Question

1. Alice made the conjecture below.(a + b)2 = a + b2OWhich values of a and b are not counterexamples to the conjecture?a = 1, b = 1a = 0, b = 1aa = -1, b = 1a = -1, b = 2

Answer 1

the expression is

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

substitute a=1 and b=1

(1+1)^2 = 1+1^2

4=2

that is not possible. so these are the values of a and b that is not counterpart example to the conjecture.

now substitute a=0, b=1

(0+1)^2 = 0+1^2

1=1

so this is true for the above expression.

now for a=-1, b=1

(-1+1)^2 = -1+1^2

0=0

this is true.

now for a=-1, b=2

(-1+2)^2 = -1+2^2

1=3

that is not possible

so a=1, b=1 and a=1,b=2 is the values that not counterpart example to the conjecture.