Question

Find two consecutive odd +ve integers sum of whose squares is 970

Answer 1

Hey

Here is your answer,

Let one of the number be x then other number is x + 2.

Then according to question,

x2+x+22=970
⇒x2+x2+4x+4=970
⇒2x2+4x-966=0
⇒x2+2x-483=0
⇒x2+23x-21x-483=0
⇒x(x+23)-21(x+23)=0
⇒(x-21)(x+23)=0

⇒x-21=0 or x+23=0

⇒x=21 or x=-23

Since, x being an odd positive integer,
Therefore, x = 21.

Then another number will be
x+2=21+2=23

Thus, the two consecutive odd positive integers are 21 and 23.

Hope it helps 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).