Question

Find the 2 consecutive odd integerwhose sum of square is 202

Answer 1

Answer:

let the first no. be x and second no. be x+2

ATO

x^2+(x+2)^2=202

x^2+x^2+4+4x=202

2x^2+4x-198=0

by formula method

-4+-√(4)^2-4(2)(-198)÷4

-4+-√16+1584÷4

-4+40÷4

x=9

-4-40÷4

x=-11

if we put the value of x=9 we get 202

hopefully it 's correct

Answer 2

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.