Answer
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Hint: Consecutive integers are those integers that occur one after other. Like 1 and 2 are consecutive integers. So, we should assume one integer in terms of another integer.
As we know that consecutive integers one after another or in other words pairs of integers which have a difference equal to 1 are known as consecutive. Like if x is one integer then the next consecutive integer will be equal to x + 1.
Complete step-by-step answer:
Now we had to assume two integers and find the sum of their squares and equate that with 365.
So, let the first integer is equal to n.
So, the second integer that is consecutive to n will be equal to n + 1.
According to the question.
\[{n^2} + {\left( {n + 1} \right)^2} = 365\] -----(1)
So, we had to solve equation 1 to find the value of n.
Now as we know that if a and b are two integers then \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
So, equation 1 becomes,
\[{n^2} + {n^2} + 1 + 2n = 365\]
\[2{n^2} + 2n = 365 - 1 = 364\]
Now dividing both sides of the above equation by 2 and then taking RHS of the equation to LHS of the equation. We get,
\[{n^2} + n - 182 = 0\]
Now to solve the above quadratic equation, we had to split the middle term of the above equation.
\[
{n^2} + 14n - 13n - 182 = 0 \\
n\left( {n + 14} \right) - 13\left( {n + 14} \right) = 0 \\
\]
Now taking factor (n + 14) common from the above equation. We get,
\[\left( {n - 13} \right)\left( {n + 14} \right) = 0\]
So, n = 13 or – 14.
Now, if n = 13 then n + 1 = 14
And, if n = – 14 then n + 1 = – 13
Hence, two consecutive integers whose sum of squares is equal to 365 can be 13, 14 or –14, – 13.
Note:- We should remember that if it is stated that the numbers are integers then they can also be negative because integers are negative also. So, we should not neglect negative numbers. And if we assume two consecutive numbers as n and m then it will be difficult to get the value of n and m because we have only one equation. So, we should assume two numbers such that one depends on another.
As we know that consecutive integers one after another or in other words pairs of integers which have a difference equal to 1 are known as consecutive. Like if x is one integer then the next consecutive integer will be equal to x + 1.
Complete step-by-step answer:
Now we had to assume two integers and find the sum of their squares and equate that with 365.
So, let the first integer is equal to n.
So, the second integer that is consecutive to n will be equal to n + 1.
According to the question.
\[{n^2} + {\left( {n + 1} \right)^2} = 365\] -----(1)
So, we had to solve equation 1 to find the value of n.
Now as we know that if a and b are two integers then \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
So, equation 1 becomes,
\[{n^2} + {n^2} + 1 + 2n = 365\]
\[2{n^2} + 2n = 365 - 1 = 364\]
Now dividing both sides of the above equation by 2 and then taking RHS of the equation to LHS of the equation. We get,
\[{n^2} + n - 182 = 0\]
Now to solve the above quadratic equation, we had to split the middle term of the above equation.
\[
{n^2} + 14n - 13n - 182 = 0 \\
n\left( {n + 14} \right) - 13\left( {n + 14} \right) = 0 \\
\]
Now taking factor (n + 14) common from the above equation. We get,
\[\left( {n - 13} \right)\left( {n + 14} \right) = 0\]
So, n = 13 or – 14.
Now, if n = 13 then n + 1 = 14
And, if n = – 14 then n + 1 = – 13
Hence, two consecutive integers whose sum of squares is equal to 365 can be 13, 14 or –14, – 13.
Note:- We should remember that if it is stated that the numbers are integers then they can also be negative because integers are negative also. So, we should not neglect negative numbers. And if we assume two consecutive numbers as n and m then it will be difficult to get the value of n and m because we have only one equation. So, we should assume two numbers such that one depends on another.
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