Sure, let's show that (a + b)^2 is equal to a + 2ab + b^2 when a = 2 and b = 1.
Given:
a = 2
b = 1
Now, let's calculate (a + b)^2:
(a + b)^2 = (2 + 1)^2
= 3^2
= 9
Next, let's calculate a + 2ab + b^2:
a + 2ab + b^2 = 2 + 2(2)(1) + 1^2
= 2 + 4 + 1
= 7
Since (a + b)^2 is equal to a + 2ab + b^2 when a = 2 and b = 1, we can see that 9 = 7. However, this is not correct.
Let's reevaluate the calculation:
(a + b)^2 = (2 + 1)^2
= 3^2
= 9
a + 2ab + b^2 = 2 + 2(2)(1) + 1^2
= 2 + 4 + 1
= 7
Upon reviewing the calculations, it appears that there was an error in the initial claim. The equation (a + b)^2 = a + 2ab + b^2 is not true for the given values of a = 2 and b = 1. The correct result is (a + b)^2 = 9, while a + 2ab + b^2 = 7.
Answers & Comments
Answer:
Sure, let's show that (a + b)^2 is equal to a + 2ab + b^2 when a = 2 and b = 1.
Given:
a = 2
b = 1
Now, let's calculate (a + b)^2:
(a + b)^2 = (2 + 1)^2
= 3^2
= 9
Next, let's calculate a + 2ab + b^2:
a + 2ab + b^2 = 2 + 2(2)(1) + 1^2
= 2 + 4 + 1
= 7
Since (a + b)^2 is equal to a + 2ab + b^2 when a = 2 and b = 1, we can see that 9 = 7. However, this is not correct.
Let's reevaluate the calculation:
(a + b)^2 = (2 + 1)^2
= 3^2
= 9
a + 2ab + b^2 = 2 + 2(2)(1) + 1^2
= 2 + 4 + 1
= 7
Upon reviewing the calculations, it appears that there was an error in the initial claim. The equation (a + b)^2 = a + 2ab + b^2 is not true for the given values of a = 2 and b = 1. The correct result is (a + b)^2 = 9, while a + 2ab + b^2 = 7.