Question

Find 2 consecutive integers such that the sum of their squares is 421.

Answer 1

Let x be an integer. The next consecutive integer would be "x+1". Now, we want the sum of their squares to be 421:
\(x^2+(x+1)^2=421\)

Answer 2

Now all ya gotta do is solve that ^^^ equation...

Answer 3

x^2+(x+1)^2=421

x^2 + ( x + 1) * (x + 1) = 421


@Garrett1234 --> ( x + 1) * (x + 1) =

x*( x + 1) + 1( x + 1) = ?

You will have four terms when you finish multiplying.

Answer 4

So here, x*( x + 1) + 1( x + 1) = ?, you distribute the x correct? I'm a little confused. @Directrix

Answer 5

Yes, @Garrett1234 I'll structure it a bit to help. Hold on.

Answer 6

You distribute the x in front of (x+1) and "distribute" the 1 in front of (x+1) but, since it's just a 1 it's not necessary lol. So yes.
If you know any identities, you can simply use \((a+b)^2=a^2+2ab+b^2\)

Answer 7

@Garrett1234

Picking up where we left off, we had x*( x + 1) + 1( x + 1) = ?

I am going to use the DRAW feature to show how to do the multiplication.

You get to see how that feature works.