Answer
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Hint: To attempt this question knowledge about the odd positive integer is must, one can take 2n + 1 and 2n + 3 as 2 positive odd integers and to find the value of n use the method of factorization, using this information will help you to approach the solution.
Complete step-by-step answer:
Let the two consecutive odd positive integers be (2n+1) and (2n + 3) where n is always greater than 0 i.e, n > 0
Now according to the question, the sum of the squares of (2n+1) and (2n + 3) should be equal to 970
Which can be written as; ${\left( {2n + 1} \right)^2} + {\left( {2n + 3} \right)^2} = 970$
By solving the above equation, we will get the value of n
So, using the algebraic formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the above equation we get
$4{n^2} + 1 + 4n + 4{n^2} + 9 + 12n = 970$
$ \Rightarrow $$8{n^2} + 16n + 10 = 970$
$ \Rightarrow $${n^2} + 2n - 120 = 0$
Now to find the value of n using the method of factorization we get
${n^2} + \left( {12 - 10} \right)n - 120 = 0$
$ \Rightarrow $${n^2} + 12n - 10n - 120 = 0$
$ \Rightarrow $$n\left( {n + 12} \right) - 10\left( {n + 12} \right) = 0$
$ \Rightarrow $$\left( {n - 10} \right)\left( {n + 12} \right) = 0$
Now for (n – 10) we get n = 10
For (n + 12) we get n = – 12
Since the value of n can’t be negative therefore, n = 10
Now substituting the value of n in both the odd positive integers i.e. (2n+1) and (2n + 3) we get
21, 23
Therefore, 21 and 23 are the two consecutive odd positive integers, the sum of whose squares is 970.
Note: The trick behind these types of question is to use the concept of even numbers we know that when an even number is added with an odd number it gives an odd number so as 2n is denotes as even numbers so when we add any odd number with 2n we get an odd number but since in the above question it was mentioned that the two odd positive integers are consecutive that means there is only difference of 2 between the two positive odd integers so this was the reason we took (2n + 1) and (2n + 3) as the two consecutive odd numbers where as it was also mentioned that the odd numbers are positive which gave us idea that n > 0.
Complete step-by-step answer:
Let the two consecutive odd positive integers be (2n+1) and (2n + 3) where n is always greater than 0 i.e, n > 0
Now according to the question, the sum of the squares of (2n+1) and (2n + 3) should be equal to 970
Which can be written as; ${\left( {2n + 1} \right)^2} + {\left( {2n + 3} \right)^2} = 970$
By solving the above equation, we will get the value of n
So, using the algebraic formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the above equation we get
$4{n^2} + 1 + 4n + 4{n^2} + 9 + 12n = 970$
$ \Rightarrow $$8{n^2} + 16n + 10 = 970$
$ \Rightarrow $${n^2} + 2n - 120 = 0$
Now to find the value of n using the method of factorization we get
${n^2} + \left( {12 - 10} \right)n - 120 = 0$
$ \Rightarrow $${n^2} + 12n - 10n - 120 = 0$
$ \Rightarrow $$n\left( {n + 12} \right) - 10\left( {n + 12} \right) = 0$
$ \Rightarrow $$\left( {n - 10} \right)\left( {n + 12} \right) = 0$
Now for (n – 10) we get n = 10
For (n + 12) we get n = – 12
Since the value of n can’t be negative therefore, n = 10
Now substituting the value of n in both the odd positive integers i.e. (2n+1) and (2n + 3) we get
21, 23
Therefore, 21 and 23 are the two consecutive odd positive integers, the sum of whose squares is 970.
Note: The trick behind these types of question is to use the concept of even numbers we know that when an even number is added with an odd number it gives an odd number so as 2n is denotes as even numbers so when we add any odd number with 2n we get an odd number but since in the above question it was mentioned that the two odd positive integers are consecutive that means there is only difference of 2 between the two positive odd integers so this was the reason we took (2n + 1) and (2n + 3) as the two consecutive odd numbers where as it was also mentioned that the odd numbers are positive which gave us idea that n > 0.
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