Question

the sum of squares of two consecutive odd integers is 290 then find those integers​

Answer 1

$\huge\sf\mathbb\color{yellow}\underline{\colorbox{black}{ solution....༊}}$

$\orange{ʟᴇᴛ \: x \: ᴀɴ \: ᴏᴅᴅ \: ᴘᴏsɪᴛɪᴠᴇ \: ɪɴᴛᴇɢᴇʀ }$

$\red{ᴀᴄᴄᴏʀᴅɪɴɢ \: ᴛᴏ \: ǫᴜᴇsᴛɪᴏɴ }$

$\bold\purple{ \: {x}^{2} + {(x + 2)}^{2} =290\: }$

$\purple{2 {x}^{2} + 4x - 286 = 0 }$

$\purple{ {x}^{2} + 2x - 143 = 0}$

$\purple{ {x}^{2} + 13x - 11x - 143 = 0}$

$\purple{ \:(x+13)(x−11)=0 \: }$

$\purple{ \: x=11 as \: x \: is \: positive\: }$

$\mathfrak\red{ \: \: \:Hence \: required \: integers \: are \:  11  \: and  \: 13</p><p>}$

Answer 2

$\huge\sf\mathbb\color{yellow}\underline{\colorbox{black}{ solution....༊}}$

$\orange{ʟᴇᴛ \: x \: ᴀɴ \: ᴏᴅᴅ \: ᴘᴏsɪᴛɪᴠᴇ \: ɪɴᴛᴇɢᴇʀ }$

$\red{ᴀᴄᴄᴏʀᴅɪɴɢ \: ᴛᴏ \: ǫᴜᴇsᴛɪᴏɴ }$

$\bold\purple{ \: {x}^{2} + {(x + 2)}^{2} =290\: }$

$\purple{2 {x}^{2} + 4x - 286 = 0 }$

$\purple{ {x}^{2} + 2x - 143 = 0}$

$\purple{ {x}^{2} + 13x - 11x - 143 = 0}$

$\purple{ \:(x+13)(x−11)=0 \: }$

$\purple{ \: x=11 as \: x \: is \: positive\: }$

$\mathfrak\red{ \: \: \:Hence \: required \: integers \: are \:  11  \: and  \: 13</p><p>}$