if 2 is the root of quadratic equation 3xsqare +px-8=0 and 4xsquare -2px+k=0. find the value of k
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Put x= 2 in given eqn
3+(2)sq -8=0
12+2p-8=0
2p+4=0
p=-2
Now put In another eqn
4(2)sq -2(-2)(2)+k=0
16+8+k=0
K= -24 Ans
honeyaman4561:
2p +4 aaga
Answered by
0
2 ɪs ᴛʜᴇ ʀᴏᴏᴛ ᴏғ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ 3x² + ᴘx - 8 = 0………….(1)
& 4x² - 2ᴘx + ᴋ = 0 ʜᴀs ᴇǫᴜᴀʟ ʀᴏᴏᴛs ………….(2)
ᴏɴ ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ɢɪᴠᴇɴ ʀᴏᴏᴛ ɪ.ᴇ x = 2 ɪɴ ᴇǫ 1 .
3x² + ᴘx - 8 = 0
3(2)² + ᴘ(2) − 8 = 0
3 × 4 + 2ᴘ - 8 = 0
12 + 2ᴘ − 8 = 0
4 + 2ᴘ = 0
2ᴘ = - 4
ᴘ = - 4/ 2 = -2
ᴘ = - 2
ʜᴇɴᴄᴇ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴘ ɪs - 2.
ᴏɴ ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴘ = - 2 ɪɴ ᴇǫ 2,
4x² - 2ᴘx + ᴋ = 0
4x² - 2(- 2)x + ᴋ = 0
4x² + 4x + ᴋ = 0
ᴏɴ ᴄᴏᴍᴘᴀʀɪɴɢ ᴛʜᴇ ɢɪᴠᴇɴ ᴇǫᴜᴀᴛɪᴏɴ ᴡɪᴛʜ ᴀx² + ʙx + ᴄ = 0
ʜᴇʀᴇ, ᴀ = 4, ʙ = 4 , ᴀɴᴅ ᴄ = ᴋ
ᴅ(ᴅɪsᴄʀɪᴍɪɴᴀɴᴛ) = ʙ² – 4ᴀᴄ
ɢɪᴠᴇɴ : ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ʜᴀs ᴇǫᴜᴀʟ ʀᴏᴏᴛs ɪ.ᴇ ᴅ = 0
ʙ² – 4ᴀᴄ = 0
4² – 4(4)(ᴋ) = 0
16 – 16ᴋ = 0
16 = 16ᴋ
ᴋ = 16/16 = 1
ᴋ = 1
ʜᴇɴᴄᴇ, ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴋ ɪs 1 .
★★ ɴᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ʀᴏᴏᴛs
ɪғ ᴅ = 0 ʀᴏᴏᴛs ᴀʀᴇ ʀᴇᴀʟ ᴀɴᴅ ᴇǫᴜᴀʟ
ɪғ ᴅ > 0 ʀᴏᴏᴛs ᴀʀᴇ ʀᴇᴀʟ ᴀɴᴅ ᴅɪsᴛɪɴᴄᴛ
ɪғ ᴅ < 0 ɴᴏ ʀᴇᴀʟ ʀᴏᴏᴛs
& 4x² - 2ᴘx + ᴋ = 0 ʜᴀs ᴇǫᴜᴀʟ ʀᴏᴏᴛs ………….(2)
ᴏɴ ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ɢɪᴠᴇɴ ʀᴏᴏᴛ ɪ.ᴇ x = 2 ɪɴ ᴇǫ 1 .
3x² + ᴘx - 8 = 0
3(2)² + ᴘ(2) − 8 = 0
3 × 4 + 2ᴘ - 8 = 0
12 + 2ᴘ − 8 = 0
4 + 2ᴘ = 0
2ᴘ = - 4
ᴘ = - 4/ 2 = -2
ᴘ = - 2
ʜᴇɴᴄᴇ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴘ ɪs - 2.
ᴏɴ ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴘ = - 2 ɪɴ ᴇǫ 2,
4x² - 2ᴘx + ᴋ = 0
4x² - 2(- 2)x + ᴋ = 0
4x² + 4x + ᴋ = 0
ᴏɴ ᴄᴏᴍᴘᴀʀɪɴɢ ᴛʜᴇ ɢɪᴠᴇɴ ᴇǫᴜᴀᴛɪᴏɴ ᴡɪᴛʜ ᴀx² + ʙx + ᴄ = 0
ʜᴇʀᴇ, ᴀ = 4, ʙ = 4 , ᴀɴᴅ ᴄ = ᴋ
ᴅ(ᴅɪsᴄʀɪᴍɪɴᴀɴᴛ) = ʙ² – 4ᴀᴄ
ɢɪᴠᴇɴ : ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ʜᴀs ᴇǫᴜᴀʟ ʀᴏᴏᴛs ɪ.ᴇ ᴅ = 0
ʙ² – 4ᴀᴄ = 0
4² – 4(4)(ᴋ) = 0
16 – 16ᴋ = 0
16 = 16ᴋ
ᴋ = 16/16 = 1
ᴋ = 1
ʜᴇɴᴄᴇ, ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴋ ɪs 1 .
★★ ɴᴀᴛᴜʀᴇ ᴏғ ᴛʜᴇ ʀᴏᴏᴛs
ɪғ ᴅ = 0 ʀᴏᴏᴛs ᴀʀᴇ ʀᴇᴀʟ ᴀɴᴅ ᴇǫᴜᴀʟ
ɪғ ᴅ > 0 ʀᴏᴏᴛs ᴀʀᴇ ʀᴇᴀʟ ᴀɴᴅ ᴅɪsᴛɪɴᴄᴛ
ɪғ ᴅ < 0 ɴᴏ ʀᴇᴀʟ ʀᴏᴏᴛs
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