• Jul 12th 2008, 03:04 PM
dancingqueen9
Aaaalright, apparently I'm supposed to solve this using the quadratic formula (ps: you dont have to run through the steps of the formula with me, i'm good with that part)

Here is the word problem: The sum of the squares of two consecutive numbers is 481. Find the integers.

How do I solve this?

Thank you :)
• Jul 12th 2008, 03:44 PM
cyph1e
Let the smaller integer be $\displaystyle x$, then the consecutive integer is $\displaystyle x+1$. The sum of the squares of these numbers is 481, which leads us to the equation $\displaystyle x^2+(x+1)^2=481$.
• Jul 13th 2008, 08:53 AM
dancingqueen9
But then, won't it end up having a degree of 4? Is that still ok?
• Jul 13th 2008, 09:36 AM
dancingqueen9
Oh, never mind. lol it does work
• Jul 13th 2008, 10:00 AM
janvdl
Quote:

Originally Posted by cyph1e
Let the smaller integer be $\displaystyle x$, then the consecutive integer is $\displaystyle x+1$. The sum of the squares of these numbers is 481, which leads us to the equation $\displaystyle x^2+(x+1)^2=481$.

Quote:

Originally Posted by dancingqueen9
But then, won't it end up having a degree of 4? Is that still ok?

Just to make sure you understand, let me build on cyph1e's foundation.

$\displaystyle x^2+(x+1)^2=481$

$\displaystyle x^2+(x^2 + 2x + 1) = 481$

$\displaystyle 2x^2 + 2x + 1 =481$

$\displaystyle 2x^2 + 2x - 480 = 0$

$\displaystyle 2(x^2 + x - 240) = 0$

$\displaystyle 2(x + 16)(x - 15) = 0$
• Jul 13th 2008, 11:43 AM
cyph1e
Quote:

Originally Posted by janvdl
These numbers are not consecutive, although there squares do add up to 481. Their absolute values are consecutive though. (15 and 16)

(Unless I did something stupid again... (Wondering))

There are two solutions; $\displaystyle \left\{(15,16),(-16,-15)\right\}$. The quadratic equation gives only the smaller integer $\displaystyle x$, while the other consecutive integer of $\displaystyle x$ is $\displaystyle x+1$.
• Jul 13th 2008, 11:45 AM
janvdl
Quote:

Originally Posted by cyph1e
There are two solutions; $\displaystyle \left\{(15,16),(-16,-15)\right\}$. The quadratic equation gives only the smaller integer $\displaystyle x$, while the other consecutive integer of $\displaystyle x$ is $\displaystyle x+1$.

Ah, of course you are correct! I knew I forgot something simple again. (Headbang)