Question

find 2 consecutive odd positive integers sum of whose squares is 970

Answer 1

Hi Mate !!


Let the first consecutive odd integer be x
and second be ( x + 2 ) as it is given they are odd.


• Sum of square is 970

x² + ( x + 2 )² = 970

[ Using identity :- ( a + b )² = a² + 2ab + b² ]

x² + x² + 4x + 4 = 970

2x² + 4x + 4 - 970 = 0

2x² + 4x - 966 = 0

{ Dividing both side by 2 }

x² + 2x - 483 = 0


Now , by factorising above equation !!


x² + 23x - 21x - 483 = 0

x ( x + 23 ) - 21 ( x + 23 ) = 0

( x + 23 ) ( x - 21 ) = 0


° ( x + 23 ) = 0

x = ( - 23 ) ..... [ It is neglected bcoz , it is given in the question that the nos. are positive ]


° ( x - 21 ) = 0

x = 21


So, 1st no.is x = 21
and 2nd no. is ( x + 2 ) = 23

And hence , 21 and 23 are the consecutive positive odd integer having sum of the square as 970 !!

Answer 2

ʟᴇᴛ ᴛʜᴇ ᴏɴᴇ ᴄᴏɴsᴇᴄᴜᴛɪᴠᴇ ᴏᴅᴅ ɪɴᴛᴇɢᴇʀ ʙᴇ (x + 1) ᴀɴᴅ ᴏᴛʜᴇʀ ᴄᴏɴsᴇᴄᴜᴛɪᴠᴇ ᴏᴅᴅ ɪɴᴛᴇɢᴇʀ ʙᴇ (x + 3).

ᴀ .ᴛ.ǫ

(x + 1)² + (x + 3)² = 970

x² + 1 + 2x + x² + 9 + 6x = 970

[(ᴀ + ʙ)² = ᴀ² + ʙ² + 2ᴀʙ ]

2x² + 8x + 10 = 970

2x² + 8x + 10 - 970 = 0

2x² + 8x - 960 = 0

2(x² + 4x - 480) = 0

x² + 4x - 480 = 0

x² + 24x - 20x  - 480 = 0

[ʙʏ ᴍɪᴅᴅʟᴇ ᴛᴇʀᴍ sᴘʟɪᴛᴛɪɴɢ]

x(x + 24) - 20(x + 24) = 0

(x - 20) (x + 24) = 0

(x - 20)  = 0  ᴏʀ (x + 24) = 0

x = 20 ᴏʀ x = - 24
sɪɴᴄᴇ, x ɪs ᴀ ᴘᴏsɪᴛɪᴠᴇ ɪɴᴛᴇɢᴇʀ, sᴏ x ≠ - 24

ᴛʜᴇʀᴇғᴏʀᴇ , x = 20

ғɪʀsᴛ ᴏᴅᴅ ɴᴜᴍʙᴇʀ (x + 1) = 20 + 1 = 21

sᴇᴄᴏɴᴅ ᴏᴅᴅ ɴᴜᴍʙᴇʀ (x + 3) = 20 + 3 = 23

ʜᴇɴᴄᴇ, ᴛʜᴇ ᴛᴡᴏ ᴏᴅᴅ ᴘᴏsɪᴛɪᴠᴇ ɴᴜᴍʙᴇʀs ʙᴇ (21, 23).