Math, asked by BrainlyHelper, 1 year ago

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

Answered by nikitasingh79
149

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.

HOPE THIS ANSWER WILL HELP YOU...


jarpana2003: really nice answer
jarpana2003: pls mark this as brainlist
Answered by mysticd
44

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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