Question

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2 points
38. If x2 + k(4x + k – 1) + 2 = 0 has equal roots, then k =
(a)
(b) ,-1
(c)
3 1
(d)
ilm
1
2 3
а
Ob
O
C
Od
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Answer 1

Answer:

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.