The values of k, for which the equation 
possess atleast one positive root, are 
. Find value of 
.
Text Solution
Verified by Experts1
(1)
Sol. Case-I : Both the roots are positive
X 2 + 2 (K – 1) x + (K + 5) = 0
(i) D 
0
⇒ 4(K – 1) 2 – 4 (K + 5) 
0
⇒ (K + 1) (K – 4) 
0
(ii) f(0) > 0
⇒ K + 5 > 0
⇒ K > – 5
(iii) 
> 0
⇒ 
> 0
⇒ K < 1
K ∈ (– ∞ , – 1]... (i)
Case-II: One root is +ve and other root is –ve
f(0) < 0
⇒ k + 5 < 0
⇒ K < – 5 ... (ii)
Case-III: One root is zero and other is +ve
f(0) = 0 & 
> 0
⇒ K = – 5 ... (iii)
Union of all the three cases give K ∈ (– ∞ , – 1] = (– ∞ , – b] ⇒ b = 1.
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