If x²+k(4x+k-1)+2=0 has equal roots, then k =
(a)-2/3, 1
(b)2/3, -1
(c)3/2, 1/3
(d)-3/2, -1/3
Answers
SOLUTION :
Option (b) is correct : ⅔ , - 1
Given : x² + k(4x + k - 1) + 2 = 0
x² + 4kx + k² - k + 2 = 0
On comparing the given equation with ax² + bx + c = 0
Here, a = 1 , b = 4k , c = k² - k + 2
D(discriminant) = b² – 4ac
D = (4k)² - 4 × 1 × (k² - k + 2)
D = 16k² - 4k² + 4k - 8
D = 12k² + 4k - 8
D = 4(3k² + k - 2)
D = 0 (equal roots given)
4(3k² + k - 2) = 0
(3k² + k - 2) = 0
3k² + 3k - 2k - 2 = 0
[By middle term splitting]
3k (k + 1) - 2(k + 1) = 0
(3k - 2)(k + 1) = 0
(3k - 2) = 0 (k + 1) = 0
3k = 2 or k = - 1
k = ⅔ or k = - 1
Hence, the value of k is ⅔ & - 1.
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Solution :
Given Quadratic equation:
x²+k(4x+k-1)+2=0
=> x² + 4kx + k² - k + 2 = 0
=> x² + 4kx + ( k² - k + 2 )=0
Compare above equation
with ax²+bx+c = 0 ,we get
a = 1 , b = 4k , c = k²-k+2
Now ,
Discreminant (D)=0
[ Given , roots are equal ]
=> b² - 4ac = 0
=> (4k )² -4×1×(k²-k+2)=0
=> 16k² - 4(k²-k+2)=0
=> 4[ 4k² -(k²-k+2)] = 0
=> 4k² - ( k²-k+2) = 0
=> 4k² - k² + k - 2 = 0
=> 3k² + k - 2 = 0
=> 3k² + 3k - 2k - 2 = 0
=> 3k( k + 1 ) -2( k + 1 ) = 0
=> ( k +1 )( 3k -2 ) = 0
k + 1 = 0 or 3k - 2 = 0
=> k = -1 or k = 2/3
Therefore ,
Option ( b ) is correct.
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