Location of Roots when Compared with a constant 'k'
If both the r...
Question
If both the roots of the equation x2+2(k+1)x+9k−5=0 are negative, then the least positive integral value of k is
A
1.00
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B
1
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C
1.0
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Solution
Let f(x)=x2+2(k+1)x+9k−5.
Roots of x2+2(k+1)x+9k−5=0 are α,β
Given: α,β<0
Conditions that needs to be satisfied are: (i)D≥0⇒4(k+1)2−4(9k−5)≥0⇒k2−7k+6≥0 ⇒(k−1)(k−6)≥0⇒k∈(−∞,1]∪[6,∞)⋯(1)
(ii)−b2a<0⇒−(k+1)<0⇒k>−1⋯(2)
(iii)f(0)>0⇒9k−5>0⇒k>59⋯(3)
From (1),(2) and (3), we get:
k∈(59,1]∪[6,∞)
Hence, the least positve integral value of k is 1.