Two consecutive numbers whose squares add up to 685

Question 790617: The sum of the squares of two consecutive integers is 685. Find the integers.

I tried solving this myself and got 324 and 325 which is not the right answer, I'm not very good with word problems. Thank you in advance!

(Scroll Down for Answer!)

Did you know that Algebra.Com has hundreds of free volunteer tutors who help people with math homework? Anyone can ask a math question, and most questions get answers!

OR get immediate PAID help on:

Go!!!

Found 2 solutions by rothauserc, stanbon:Answer by rothauserc(4570) About Me  (Show Source):

You can put this solution on YOUR website!

we are given the following

x^2 +(x+1)^2 = 685

x^2 +x^2+2x+1 = 685

2x^2 +2x = 684

x^2 +x = 342 and

x^2 +x -342 = 0

use quadratic equation to solve for x

x = (-1 + or - square root(1 - 4*1*(-342)) / 2

x = (-1 + or - 37) / 2

the problem does not state if our consecutive integers are positive or negative

x = 18 or -19

18^2 + 19^2 = 685

324 + 361 = 685

685 = 685

now

-19^2 + -18^2 = 685

361 + 325 = 685

so our integers are 18, 19, -19, -18

Answer by stanbon(75874) About Me  (Show Source):

You can put this solution on YOUR website!

The sum of the squares of two consecutive integers is 685. Find the integers.

1st: x

2nd: x+1

Equation:

x^2 + (x+1)^2 = 685

x^2 + x^2 + 2x + 1 = 685

Step-by-step explanation:

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )

sine

step-by-step explanation: