Question & Answer
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 Verified Verified
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.