Question

Find two positive numbers whose difference is 8 and the sum of whose squares is 914.

Give the answers in ascending order

Answer 1

i just need the equations

Answer 2

would it be x-y=8 for the first one?

Answer 3

Yes, and x^2+y^2=914 for the second.

Answer 4

thank you, i tried that before, but i didn't get the right answer

Answer 5

can i add y^2 + y^2, and it'll be y^4?

Answer 6

i'm getting y^2 +y^2=914-64

Answer 7

but then i'm not sure what to do after; do i add the y's or do i take the square of 850?

Answer 8

If you do y^2+y^2, it will be 2y^2
Then 2y^2=850,
then divide by 2--- y^2=425
But that is not a perfect square. Hole on. Let me try something.

Answer 9

ok

Answer 10

x-y=8 ----> -y=8-x---->y=-8+x
k=x^2+y^2
k=x^2+(-8+x)^2
k=x^2+64-16x+x^2
k=2x^2-16x+64
Next we factor. Before I continue, did you understand that part?

Answer 11

yes

Answer 12

so it'll be 2(x -8)(x-8)

Answer 13

and then x-8=0
x=8 and same thing for the other x?

Answer 14

Yes, that will be right, but now, I'm trying to figure how we will find what y is.

Answer 15

ok

Answer 16

Yay I got the answer. So I did something wrong. Instead of
k=2x^2-16x+64 it should have been
914=2x^2-16x+64
Then subtract 914 from both sides.
0=2x^2-16x-850
Then divide the right side by 2
0=x^2-8x-425
Factor
0=(x-25)(x+17)
Then solve
x=25 and x=17
25-17=9
625+289=914
So the numbers are 25 and 17.
That took awhile lol

Answer 17

LOL yes that's right! Thank you so much!

Answer 18

Np. It would've bugged me all day if I didn't figure it out. :)

Answer 19

i'm sure it would have! lol

Answer 20

are you also good at logs?

Answer 21

Yes. Do you have a log question?

Answer 22

yes i do, i'll post it now :)