Question

Find 3 positive consecutive integers such that the square of the first, added to the last, is 8.

Answer 1

Say these are the consecutive integers: n,n+1,n+2

Now it says the square of the first+last=8
So we have
\[n^2+(n+2)=8\]

Answer 2

See if you can solve this for n

Answer 3

3 integers x, x+1, x+2

then x^2 + (x+2)^2 = 8

expand and simplify

x^2 + x+2 = 8

solve the quadratic

x^2 + x - 6 = 0

find the positive solution

Answer 4

oops omit the line x^2 + (x+2)^ = 8.....

Answer 5

im really confused can you draw it out?

Answer 6

Which part is confusing you?

Answer 7

Is it the 3 consecutive integers being n,n+1,n+2?

Answer 8

or is it the square of the first added to the last is 8 confusing you?

Answer 9

\[n^2+(n+2)=8\]

n is the first
n+2 is the last

Answer 10

im going to try and find n .

Answer 11

k

Answer 12

So n= -3 and n= 2

Answer 13

so i use 2

Answer 14

First step to solve for n subtract 8 on both sides:
\[n^2+n+2-8=0\]
\[n^2+n-6=0\]
\[(n+3)(n-2)=0\]
Omg you are totally correct!
go you brendahidalgo! :)

Answer 15

So but it does say positive

Answer 16

So n=2 is the positive answer of the two you have there

Answer 17

So n+1=?

Answer 18

yes it says three positive consecutive numbers

Answer 19

so the numbers are 2, 3, and 4? because 2 squared is 4, and when you add 4 it is eight.

Answer 20

yep 2,3,4 :)

Answer 21

thank you so much :)

Answer 22

Very good job brendahidalgo!