Q.
If the equation x2+2(k+1)x+9k−5=0 has only negative roots, then
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Complex Numbers and Quadratic Equations
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Solution:
Let f(x)=x2+2(k+1)x+9k−5.
Let α,β be the roots of f(x)=0.
The equation f(x)=0 will have both negative roots if and only if
(i) Disc. ≥0
(ii) α+β<0 and
(iii) f(0)>0
Now, discriminant ≥0 ⇒4(k+1)2−36k+20≥0 ⇒k2−7k+6≥0 ⇒(k−1)(k−6)≥0 ⇒k≤1
or k≥6(1) (α+β)<0 ⇒−2(k+1)<0 ⇒k+1>0 ⇒k>−1(−2)
and aβ>0 ⇒9k−5>0 ⇒k>95(3)
From Eqs (1), (2) and (3) we get k≥6.