Question 221401
Find two consecutive positive integers if the sum of their squares is 61.


Step 1.  Let n be a positive integer.


Step 2. Let n+1 be the next positive and consecutive number.


Step 3.  Let {{{n^2+(n+1)^2=61}}} since the sum of their squares is 61.


Step 4.  Solving yields the following steps


{{{n^2+n^2+2n+1=61}}}


{{{2n^2+2n+1=61}}}


Subtract 61 from both sides of the equation to get a quadratic equation.


{{{2n^2+2n+1-61=61-61}}}


{{{2n^2+2n-60}}}


Step 5.  To solve, use the quadratic formula given as


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


where a=2, b=2 and c=-60


*[invoke quadratic "x", 2, 2, -60 ]


Select the positive solution n=5 then n+1=6 and {{{5^2+6^2=61}}}...a true statement.


Step 6.  ANSWER:  The consecutive positive integers are 5 and 6


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J