Question

The sum of squares of two consecutive odd integers is 202

Answer 1

Let x and (x+2) be two consecutive odd natural numbers.

A/C,

$\longrightarrow$$\sf\green {{x}^{2} + {(x + 2)}^{2} = 202}$

$\longrightarrow$$\sf \green{{x}^{2} + {x}^{2} + 4x + 4 = 202}$

$\longrightarrow$$\sf\green {{2x}^{2} + 4x - 198 = 0}$

$\longrightarrow$$\sf \green{{x}^{2} + 2x - 99 = 0}$

$\longrightarrow$$\sf\green{(x - 9)(x + 11) = 0}$

$\longrightarrow$$\sf\green{x = \: or \: x = - 11}$

As x is a natural number, x ≠ -11 ,so x = 9

Hence, the two consecutive odd natural numbers are 9 and 11.