What is the answer to this geometry problem?

Question

Complete the conjecture based on the pattern you observe in the specific cases.

Conjecture: For any two numbers a and b, the product of (a+b) and (a-b) is always equal to ?.

(2+1)x(2-1)=3=2^2-1^2

(3+2)x(3-2)=5=3^2-2^2

(4+2)x(4-2)=12=4^2-2^2

(6+3)x(6-3)=27=6^2-3^2

Please explain how you did this. If you don't already know 5^2, for example means five to the second power.

Answer 1

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

so in first case

(2+1)(2-1)=2^2-1^2=4-1=3

same for others also

ok

Answer 2

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

(2+1)x(2-1) = (3)(1) = 3

Or... (2+1)x(2-1) = 2^2 - 1^2 = 4 - 1 = 3  (same as above)

(3+2)x(3-2)= (5)(1) = 5

Or....  = 3^2 - 2^2 = 9 - 4 = 5

I'll leave the rest up to you

Answer 3

They are basically trying to get you to see a pattern and generalize from there.  The pattern is (first number)^2 - (second number)^2

As stated by others, the generalized form of this is simply a^2 -b^2.

Answer 4

it's always the first number squared minus the second number squared.

Explaining how I did this - I looked at the pattern and it's always the first number squared minus the second number squared.

P.S.  use '*' to mean multiply, not x when typing on the computer.

_/

Answer 5

it always comes out to be a squared minus b squared

Answer 6

I don't know what you mean by your answers. They are all correct. But (a + b)(a - b) always equals a² - b².

Answer 7

(a+b)(a-b) = (a²)-(b²)

Looks like the distributive property.

Answer 8

This is called the difference between two squares and is a famous problem in Algebra.

(a^2-b^2) = Difference between two squares= (a+b)(a-b)

Notice the middle term cancels when you use the distributive property on (a+b)(a-b) = a^2-ab+ba-b^2.

-ab+ba= -ab+ab= 0 and you are left with (a^2-b^2)