Question

Find two consecutive positive integers, the sum of whose squares is \[365\].

Answer 1

Hint: We have to find the two integers by the condition with whose squares sum is \[365\].
We have to use multiple variables to solve this problem. Using the multiply variable we convert the problem into an equation and solve it to get the required solution.
Consecutive positive integers:
Consecutive positive integers are the positive integers that follow each other in order. They have a difference of \[1\] between every two positive integers.
If we take the first positive integer m then the second positive integers is \[m + 1\] , third positive integers is \[m + 2\] and so on.

Formula used: ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step-by-step answer:
It is given that the sum of squares of two consecutive positive integers is \[365\].
Let, the first positive integer is m then the second positive integers is \[m + 1\].
Now the square of the first positive integer is \[{m^2}\]
The square of the second positive integer is \[{(m + 1)^2}\]
The sum of squares of two consecutive positive integers is \[{m^2} + {(m + 1)^2}\]
Thus, we have
\[{m^2} + {(m + 1)^2} = 365\]
Using the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we get,
\[{m^2} + {m^2} + 2m + 1 = 365\]
Simplifying the above,
\[2{m^2} + 2m + 1 - 365 = 0\]
\[2{m^2} + 2m - 364 = 0\]
Solving the quadratic equation by middle term process
\[2{m^2} + 28m - 26m - 364 = 0\]
Splitting into two terms,
\[2m(m + 14) - 26(m + 14) = 0\]
Solve for $m$ we get,
\[(m + 14)(2m - 26) = 0\]
\[m + 14 = 0{\text{ or }}(2m - 26) = 0\]
\[\begin{gathered}
  m = - 14\,{\text{or }}\dfrac{{26}}{2} \\
  m = - 14\,{\text{or }}13 \\
\end{gathered} \]
\[m = 13\] since it is given that the consecutive numbers are positive integers.
So, the second positive integer is \[13 + 1 = 14\]
Hence, two consecutive positive integers are \[13\] \[\&\] \[14\]

Additional Information:
Whole numbers:
Whole numbers are simply the numbers \[0,\,{\text{ }}1,\,{\text{ }}2,\;\;3,\;{\text{ }}4,\;{\text{ }}5,{\text{ }}6,\; \ldots ..\]

     

Integers:
Integers are like whole numbers, but they also include negative numbers but still no fractions allowed.

     

So integers can be negative{-1,-2,-3,……} , positive {1,2,3,4,……} or zero {0}.

Note: We have used the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
Quadratic equations can be solved by a middle term process.
Splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors.
Solving the two factors we will get the solution.