# Math Help - word problem - quadratics

1. ## word problem - quadratics

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

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

3. But then, won't it end up having a degree of 4? Is that still ok?

4. Oh, never mind. lol it does work

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

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

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

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

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

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

$2(x + 16)(x - 15) = 0$

6. 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... )
There are two solutions; $\left\{(15,16),(-16,-15)\right\}$. The quadratic equation gives only the smaller integer $x$, while the other consecutive integer of $x$ is $x+1$.

7. Originally Posted by cyph1e
There are two solutions; $\left\{(15,16),(-16,-15)\right\}$. The quadratic equation gives only the smaller integer $x$, while the other consecutive integer of $x$ is $x+1$.
Ah, of course you are correct! I knew I forgot something simple again.