- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Given:
The sum of the squares of two consecutive multiples of 7 is 637.
To do:
We have to find the multiples.
Solution:
Let the two consecutive multiples of 7 be $7x$ and $7x+7$.
According to the question,
$(7x)^2+(7x+7)^2=637$
$49x^2+49x^2+98x+49=637$
$49(2x^2+2x+1)=49\times13$
$2x^2+2x+1=13$
$2x^2+2x+1-13=0$
$2x^2+2x-12=0$
$2(x^2+x-6)=0$
$x^2+x-6=0$
Solving for $x$ by factorization method, we get,
$x^2+3x-2x-6=0$
$x(x+3)-2(x+3)=0$
$(x+3)(x-2)=0$
$x+3=0$ or $x-2=0$
$x=-3$ or $x=2$
The value of $x$ cannot be negative. Therefore, the value of $x$ is $2$.
$7x=7(2)=14$
$7x+7=14+7=21$
The required multiples are $14$ and $21$.
Advertisements