The sum of the SQUARES of two consecutive positive integers is 145. How do you find the numbers?

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The sum of the SQUARES of two consecutive positive integers is 145. How do you find the numbers?

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Answer 1 · 2016-01-27T11:20:53

$n²+(n+1)²=145, =n²+n²+2n+1=145, 2n²+2n=144, n²+n=72, n²+n-72=0. n=(-b+-(b²-4*a*c))/2*a,( -1+(1-4*1*-72)^0.5)/2, =(-1+(289)^0.5)/2,=(-1+17)/2=8$ . n=8, n+1=9.

Explanation:

given.

Answer 2 · 2016-01-27T11:26:17

I found $8 and 9$

Explanation:

Let us call the numbers:

$n$
and
$n+1$

we get (from our condition) that:

$(n)^2+(n+1)^2=145$

rearrange and solve for $n$ :
$n^2+n^2+2n+1-145=0$
$2n^2+2n-144=0$
use the Quadratic Formula:
$n_(1,2)=(-2+-sqrt(4+1152))/4=(-2+-34)/4$
so ve get two values:
$n_1=-9$
$n_2=8$
we chose the positive one so that our numbers will be:
$n=8$
and
$n+1=9$

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The sum of the SQUARES of two consecutive positive integers is 145. How do you find the numbers?

2 Answers

Explanation:

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