
Find two consecutive positive integers, the sum of whose squares is 365.
Answer
589.8k+ views
Hint: 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 is \[m\] then the second positive integers are \[m + 1\], third positive integers are \[m + 2\], and so on.
Formula used: ${(a + b)^2} = {a^2} + 2ab + {b^2}$
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 ${(a + b)^2} = {a^2} + 2ab + {b^2}$, we get,
\[{m^2} + {m^2} + 2m + 1 = 365\]
Simplifying as the equation, we get,
\[2{m^2} + 2m + 1 - 365 = 0\]
\[2{m^2} + 2m - 364 = 0\]
Solving the quadratic equation by middle term process
\[2{m^2} + 28m - 26m - 364 = 0\]
The above equation split into two terms,
\[2m(m + 14) - 26(m + 14) = 0\]
Taking the common terms as a root we get,
\[(m + 14)(2m - 26) = 0\]
Therefore the roots are,
\[m + 14 = 0{\text{ or }}(2m - 26) = 0\]
Hence,
\[m = - 14{\text{ or }}\dfrac{{26}}{2}\]
\[m = - 14{\text{ or }}13\]
\[m{\text{ }} = {\text{ }}13\]
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 \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6 \ldots ..\]
Integers:
Integers are like whole numbers, but they also include negative numbers but still, no fractions are allowed.
So integers can be negative \[\left\{ { - 1,{\text{ }} - 2,{\text{ }} - 3,{\text{ }} \ldots \ldots } \right\}\], positive \[\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }} \ldots \ldots } \right\}\] or zero \[\left\{ 0 \right\}\].
Note:
We have used the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
A quadratic equation can be solved by the 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.
If we take the first positive integer is \[m\] then the second positive integers are \[m + 1\], third positive integers are \[m + 2\], and so on.
Formula used: ${(a + b)^2} = {a^2} + 2ab + {b^2}$
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 ${(a + b)^2} = {a^2} + 2ab + {b^2}$, we get,
\[{m^2} + {m^2} + 2m + 1 = 365\]
Simplifying as the equation, we get,
\[2{m^2} + 2m + 1 - 365 = 0\]
\[2{m^2} + 2m - 364 = 0\]
Solving the quadratic equation by middle term process
\[2{m^2} + 28m - 26m - 364 = 0\]
The above equation split into two terms,
\[2m(m + 14) - 26(m + 14) = 0\]
Taking the common terms as a root we get,
\[(m + 14)(2m - 26) = 0\]
Therefore the roots are,
\[m + 14 = 0{\text{ or }}(2m - 26) = 0\]
Hence,
\[m = - 14{\text{ or }}\dfrac{{26}}{2}\]
\[m = - 14{\text{ or }}13\]
\[m{\text{ }} = {\text{ }}13\]
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 \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6 \ldots ..\]
Integers:
Integers are like whole numbers, but they also include negative numbers but still, no fractions are allowed.
So integers can be negative \[\left\{ { - 1,{\text{ }} - 2,{\text{ }} - 3,{\text{ }} \ldots \ldots } \right\}\], positive \[\left\{ {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }} \ldots \ldots } \right\}\] or zero \[\left\{ 0 \right\}\].
Note:
We have used the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
A quadratic equation can be solved by the 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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

