Question

Find two consecutive odd integers the sum of whose square is 130.
x+(x+2)=130
x+(x+2)-130=0
does this look right so far the square part kind of confuses me :(

Answer 1

sum of squares means, x^2 + (x+2)^2 =130

Answer 2

x^2 + (x+2)^2 =130
can u solve this ahead ?

Answer 3

is it x^2(x+2)(x+2)-130=0
and then after you foil the quantity of (x+2)^2
then you just factor right?

Answer 4

yes, but u missed a + sign in between, its x^2+(x+2)(x+2)-130=0
then foil.

Answer 5

or u can use \((x+a)^2=x^2+2ax+a^2\)

Answer 6

i would make a guess and see if it worked
i guess 9 and 11
\[9^2+11^2=202\]nope too big

Answer 7

guess i will try 7 and 9
yup, that works

Answer 8

no need to solve some silly equation for this, a little experimentation will do
the numbers are not that big, so you don't have too many choices

Answer 9

when i foiled i got x^2+(x^2+2x+2x+4)-130=0
so then i would get x^2+x^2+4x+4-130=0
2x^4+4x-126=0
then i got lost after i did this...

Answer 10

now divide the whole equation by 2, and u get a nice quadratic in x with leading co-efficient =1.

Answer 11

and which you can factor easily.

Answer 12

yeah i got thanks so much! :D

Answer 13

ok, welcome ^_^