Question

If the sum of squares of two consecutive odd natural numbers is 202 then find the numbers

Answer 1

Hey mate......☺☺☺☺



⏩Here is your answer⤵⤵⤵⤵


⏩ Let the consecutive natural numbers be x and x+1.

Then ,
${x}^{2} + ({x + 1})^{2} = 202$
${x}^{2} + {x}^{2} + 1 + 2x = 202 \\ \\ 2 {x}^{2} + 2x = 200 \\ \\ {x}^{2} + x = 100$




❤❤Hope it will help you ❤❤


✌✌✌

Answer 2

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

A/C,

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

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

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

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

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

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

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

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