Find the values of k for which the quadratic equation (3k+1)x²+2k+1x+1=0 has equal roots. Also, find the roots.
Answers
SOLUTION :
Given : (3k + 1)x² + 2(k+1)x + 1 = 0 ………(1)
On comparing the given equation with ax² + bx + c = 0
Here, a = 3k + 1 , b = 2(k+1) , c = 1
D(discriminant) = b² – 4ac
D = (2(k+1))² - 4 × (3k + 1) × 1
D = 4(k² + 1²+ 2k) - 4(3k + 1)
D = 4k² + 4 + 8k - 12k - 4
D = 4k² + 8k - 12k - 4 + 4
D = 4k² - 4k
D = 4k(k - 1)
Given : Equal roots
Therefore , D = 0
4k(k - 1) = 0
4k = 0 or (k - 1) = 0
k = 0/4 or k = 1
k = 0 or k = 1
Hence, the value of k is 0 & 1 .
On putting k = 0 in eq 1 ,
(3k + 1)x² + 2(k+1)x + 1 = 0
(3 × 0 + 1) x² + 2(0 + 1) x + 1 = 0
(0 + 1)x² + 2x + 1 = 0
x² + 2x + 1 = 0
x² + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1) = 0 or (x + 1) = 0
x = - 1 or x = - 1
Roots are - 1 & - 1
On putting k = 1 in eq 1 ,
(3k + 1)x² + 2(k+1)x + 1 = 0
(3 × 1 + 1) x² + 2(1 + 1) x + 1 = 0
(3 + 1)x² + 2× 2x + 1 = 0
4x² + 4x + 1 = 0
4x² + 2x + 2x + 1 = 0
2x(2x + 1) + 1(2x + 1) = 0
(2x + 1) = 0 or (2x + 1) = 0
2x = - 1 or 2x = - 1
x = - ½ or x = - ½
Roots are - ½ & - ½
Hence, the roots of the equation (3k + 1)x² + 2(k+1)x + 1 = 0 are - 1 & - 1 /2.
★★ NATURE OF THE ROOTS
If D = 0 roots are real and equal
If D > 0 roots are real and distinct
If D < 0 No real roots
HOPE THIS ANSWER WILL HELP YOU…
( 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