Question

Find two consecutive positive integers, sum of whose square is 365.

Answer 1

Hint: Here we assume two consecutive positive integers as two variables and use the concept that consecutive integers have a difference of 1 between them to form the first equation. Then form up an equation of sum of squares of these two numbers and equate it to 365. Use factorization method to find the numbers.
* Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is \[(x - p)(x - q) = 0\]

Complete step-by-step solution:
We know that positive integers are the integers that lie on the right hand side of 0 on the number line. Also, we know consecutive numbers mean numbers that are exactly together i.e. right next to each other.
So, consecutive positive integers can be 1,2 ; 2,3 ; 3,4 ; …..
So if we assume the first positive integer as \[x\]
Then consecutive positive integer will be \[x + 1\]
Now we know the sum of squares of two consecutive positive integers is 365
\[ \Rightarrow {x^2} + {\left( {x + 1} \right)^2} = 365\]
Use the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] to open the value in left hand side of the equation
\[ \Rightarrow {x^2} + {x^2} + 1 + 2x = 365\]
Shift all the constant values to one side
\[ \Rightarrow 2{x^2} + 2x = 365 - 1\]
\[ \Rightarrow 2{x^2} + 2x = 364\]
Cancel 2 from both sides of the equation
\[ \Rightarrow {x^2} + x = 182\]
\[ \Rightarrow {x^2} + x - 182 = 0\]
Now we use factorization method to solve for value of x
Break the coefficient of ‘x’ in such a way that their sum is equal to coefficient of ‘x’ and their product is equal to the product of coefficient of \[{x^2}\]and the constant term. We can write \[182 = 14 \times 13\]
We can write \[1 = 14 - 13\] i.e. the coefficient of x
Substitute the value of \[1 = 14 - 13\] in equation (1)
\[ \Rightarrow {x^2} + x - 182 = {x^2} + \left( {14 - 13} \right)x - 182\]
Open the bracket in RHS of the equation
\[ \Rightarrow {x^2} + x - 182 = {x^2} + 14x - 13x - 182\]
Take common terms
\[ \Rightarrow {x^2} + x - 182 = x(x + 14) - 13(x + 14)\]
Combine the factors
\[ \Rightarrow {x^2} + x - 182 = (x + 14)(x - 13)\]
Equate factors to 0
\[x + 14 = 0\] and \[x - 13 = 0\]
Shift constant values to right hand side
\[x = - 14\] and \[x = 13\]
Since we have both integers positive, we ignore the negative value
\[ \Rightarrow x = 13\]
Now we calculate the consecutive positive integer i.e. \[13 + 1 = 14\]

\[\therefore \]The two positive integers having sum of their squares 365 are 13 and 14.

Note: When shifting the values from one side of the equation to another side, always keep in mind the sign changes from positive to negative and vice versa when we shift a number from one side to another side of the equation.