Question

Find two consecutive odd positive integers, sum of whose squares is 970.

Answer 1

SOLUTION :  

Let the one consecutive odd integer be (x + 1) and other consecutive odd integer be (x + 3).

A .T.Q

(x + 1)² + (x + 3)² = 970

x² + 1 + 2x + x² + 9 + 6x = 970

[(a + b)² = a² + b² + 2ab ]

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

[By middle term splitting]

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

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

(x - 20)  = 0  or (x + 24) = 0

x = 20 or x = - 24

Since, x is a positive integer, so x ≠ - 24

Therefore , x = 20

First odd number (x + 1) = 20 + 1 = 21

Second odd number (x + 3) = 20 + 3 = 23

Hence, the two odd positive numbers be (21, 23).

HOPE THIS  ANSWER WILL HELP YOU.. .

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).