Question

Find two consecutive whole numbers such that the sum of their squares is 13

Answer 1

Putting this into algebra,
\[x^2 +(x+1)^2 = 13\], where x is the unknown
Solve for x.

Answer 2

I still can't find the two values..

Answer 3

hmm? what was you last step?

Answer 4

I got to 2x^2-13+1=0

Answer 5

Ah. I see. Note that:
\[(x+1)^2 = x^2 + 2x +1\] instead of \(x^2 + 1\)

Answer 6

I'm still having trouble finding 2 whole numbers... I got -1 + or - radical 13 What do you get when you solve it?

Answer 7

...I'm also kinda in a hurry, so if you could just tell me the answer it would really help out a lot

Answer 8

Alright, expanding the original equation,
\[x^2 + x^2+2x+1 =13 \]
\[2x^2 +2x-12=0\]
\[x^2 +x-6 =0 \]
I think you can progress from here.

Answer 9

x=3 or x=-2(Not accepted)
so x=3

Answer 10

I see now, I forgot the other x^2.. and if only 1 answer is accepted, why do I still have to provide 2 consecutive whole numbers for the question?

Answer 11

because x=2 then x+1 =3? (note:I made a mistake up there, sorry, lol)

Answer 12

lol its ok, I figured

Answer 13

it worked. thanks a lot

Answer 14

You're welcome :)