 QUESTION

# Find two consecutive positive integers, sum of whose square is 613.A. 17 and 18B. 20 and 22C. 16 and 17D. 15 and 18

Hint: Consider $x$ and $x+1$ be two consecutive numbers and they given sum of squares of these consecutive numbers are 613 i.e ${x{^2}+{(x+1){^2}}}$ =$613$. Hence solve this quadratic equation and determine the value of $x$

$\begin{array}{*{35}{l}} x{}^\text{2}\text{ }+\text{ }\left( x\text{ }+\text{ }1 \right){}^\text{2}\text{ }=\text{ }613 \\ x{}^\text{2}\text{ }+\text{ }x{}^\text{2}\text{ }+\text{ }{{1}^{2}}+\text{ }2x\text{ }=\text{ }613 \\ 2x{}^\text{2}\text{ }+\text{ }2x\text{ }=\text{ }613\text{ }\text{ }1 \\ 2{{x}^{2}}+\text{ }2x\text{ }=\text{ }612 \\ 2x{}^\text{2}\text{ }+\text{ }2x\text{ }-\text{ }612\text{ }=\text{ }0 \\ 2\text{ }\left( x{}^\text{2}\text{ }+\text{ }x\text{ }-\text{ }306 \right)\text{ }=\text{ }0 \\ x{}^\text{2}\text{ }+\text{ }x\text{ }-\text{ }306\text{ }=\text{ }0 \\ \end{array}$
Note:-Numbers that follow each other continuously in the order from smallest to largest are called consecutive numbers For Ex. 1,2,3,4,5 .Students should know the basic expansion formula of ${(a+b){^2}}$=${{a}{^2}+{b}{^2}+2ab}$ to solve these types of problems.