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Q.
If the equation $x^{2}+2(k+1) x+9 k-5=0$ has only negative roots, then
Complex Numbers and Quadratic Equations
Solution:
Let $f(x)=x^{2}+2(k+1) x+9 k-5$.
Let $\alpha, \beta$ be the roots of $f(x)=0$.
The equation $f(x)=0$ will have both negative roots if and only if
(i) Disc. $\geq 0$
(ii) $\alpha+\beta<0$ and
(iii) $f(0)>0$
Now, discriminant $\geq 0$
$\Rightarrow 4(k+1)^{2}-36 k+20 \geq 0$
$ \Rightarrow k^{2}-7 k+6 \geq 0$
$\Rightarrow (k-1)(k-6) \geq 0$
$\Rightarrow k \leq 1$
or $k \geq 6 \,\,\,\, (1)$
$(\alpha+\beta)<0$
$\Rightarrow -2(k+1)<0$
$\Rightarrow k+1>0 $
$\Rightarrow k >-1 \,\,\,\, (-2)$
and $a \beta >0$
$\Rightarrow 9 k-5 >0$
$\Rightarrow k >\frac{5}{9} \,\,\,\, (3)$
From Eqs (1), (2) and (3) we get $k \geq 6$.