Question

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

Answer 1

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 $x$

Complete step-by-step answer:
Let the two consecutive positive integers be ‘x’ and ‘x + 1’
According to the question,
\[\begin{array}{*{35}{l}}
   x{}^\text{2}\text{ }+\text{ }\left( x\text{ }+\text{ }1 \right){}^\text{2}\text{ }=\text{ }613 \\
   x{}^\text{2}\text{ }+\text{ }x{}^\text{2}\text{ }+\text{ }{{1}^{2}}+\text{ }2x\text{ }=\text{ }613 \\
   2x{}^\text{2}\text{ }+\text{ }2x\text{ }=\text{ }613\text{ }\text{ }1 \\
   2{{x}^{2}}+\text{ }2x\text{ }=\text{ }612 \\
   2x{}^\text{2}\text{ }+\text{ }2x\text{ }-\text{ }612\text{ }=\text{ }0 \\
   2\text{ }\left( x{}^\text{2}\text{ }+\text{ }x\text{ }-\text{ }306 \right)\text{ }=\text{ }0 \\
   x{}^\text{2}\text{ }+\text{ }x\text{ }-\text{ }306\text{ }=\text{ }0 \\
\end{array}\]
We have got a quadratic equation now,
x² + x - 306 = 0
x² - 18x + 17x - 306 = 0
x (x - 18) + 17 (x - 18) = 0
(x - 18) (x + 17) = 0
x = 17 or x = - 18
But x is given to be a positive integer. Therefore, x ≠ -18, x = 17.
Thus, the two consecutive positive integers are 17 and 18.

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.