Question

find the value of k for which the quadratic equation (3k+1)x2+2(k+1)x+1=0 has equal roots also find the roots

Answer 1

Here, p(x)=(3k+1)x2+2(k+1)x+1=0

Now, If  the roots are real and equal

So, D=0

i.e. b^2-4ac=0

=>(2(2k+1))^2-4(3k+1)(1)=0

=>4(4k^2+4k+1)-(12k+4)=0

=>16k^+16k+4-12k-4=0

=>(4k)(4k+1)=0

Therefore, k=0 or -1/4

Thanks!!!

Answer 2

ʜᴇʀᴇ ɪs ʏᴏᴜʀ ᴀɴsᴡᴇʀ :-


( 3ᴋ + 1 )x² + 2 ( ᴋ + 1 ) + 1 = 0


ᴄᴏᴍᴘᴀʀɪɴɢ ᴛʜᴇ ᴇǫᴜᴀᴛɪᴏɴ ᴡɪᴛʜ sᴛᴀɴᴅᴀʀᴅ ғᴏʀᴍ


ʜᴇʀᴇ


ᴀ = 3ᴋ + 1


ʙ = 2( ᴋ + 1 )


ᴄ = 1


sɪɴᴄᴇ ᴛʜᴇ ɢɪᴠᴇɴ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ʜᴀs ᴇǫᴜᴀʟ ʀᴏᴏᴛs ᴛʜᴇ ᴅɪsᴄʀɪᴍɪɴᴀɴᴛ = 0


ʙ² - 4ᴀᴄ = 0


( 3ᴋ + 1 )² - 4 ×2( ᴋ + 1 )( 1 ) = 0


=> 9ᴋ² + 6ᴋ + 1 - 8ᴋ - 8 = 0


=> 9ᴋ² - 2ᴋ - 7 = 0


=> 9ᴋ² - 9ᴋ + 7ᴋ - 7 = 0


=> 9ᴋ ( ᴋ - 1 ) + 7 ( ᴋ - 1 ) = 0


=> ( ᴋ - 1 )( 9ᴋ + 7 ) = 0


=> ᴋ - 1 = 0. ᴏʀ. 9ᴋ + 7. = 0


=> ᴋ = 1 ᴏʀ. ᴋ = -7/9