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Q. If both roots of the quadratic equation $x^2-2 k x+k^2+2 k-5=0$ are less than 4 , then exhaustive set of values of $k$ is

Complex Numbers and Quadratic Equations

Solution:

$f(x)=x^2-2 k x+k^2+2 k-5$
(i) $f (4)>0 $
$16-8 k + k ^2+2 k -5>0 $
$k ^2-6 k +11>0 $
$k \in R$
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(ii)$D \geq 0 $
$4 k^2-4\left(k^2+2 k-5\right) \geq 0 $
$-2 k+5 \geq 0$
$\therefore k \leq \frac{5}{2}$
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(iii)$\frac{-b}{2 a}<4 $
$\frac{2 k}{2}<4 $
$\therefore k<4$
Hence, from (i) $\cap$ (ii) $\cap$ (iii)
$k \in\left(-\infty, \frac{5}{2}\right]$