Question

how do I solve the following equation?
(1+2)^2=1^2+2(2)(1)+(2)^2

Answer 1

There is nothing to "solve for"

Answer 2

than simplify for?
it says Demonstrate that your polynomial identity works on numerical relationships

Answer 3

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

\(\huge\cal\ (1+2)^2-(1^2+2*(2)*(1)+(2)^2)=0\)
Step 1 :

Simplify \(\huge\ 32 - 9\)

Equation at the end of step 1 :

0 = 0

Step 2 :

Solve 0 = 0


This equation is a tautology (Something which is always true)

\(\huge\cal\ Solution: Equation~is~always~true\)

Answer 4

$$\Huge (1+2)^2=1^2+2(2)(1)+(2)^2$$ $$\Huge(a+b)^2=a^2+2(a)(b)+(b)^2$$

Answer 5

the (a+b)^2=A^2+2(a)(b)+(b)^2 is what I started with

Answer 6

The left hand side is the square of a binomial and the right hand side is called a perfect square trinomial.

Answer 7

Let a=1 and b=2

Answer 8

Right I understand that but than it said to demonstrate with numerical relationships so I picked 1 and 2

Answer 9

Yes, now simplify the right hand side and the left hand side then see if you get a true statement.

Answer 10

9 = 9

Answer 11

Is that^ true?

Answer 12

true

Answer 13

You get 9 for both sides

Answer 14

Your choice of a=1 and b=2 could be changed to any numbers and would you still get a true statement?

Answer 15

Yes

Answer 16

This is called an "identity" it is a statement that is true for every value of the variable(s).

Answer 17

Thank you!

Answer 18

Thanks for asking :)