In question number 2,
I did like this,
k+4/k−2 >0
And also i put −b/a value in the equation <0
Then used quadratic formula
64−4(k-2)(k+4) <0
And by trail and error,i got answer as 2.
Is there another way rather than intuition
In question number 2,
I did like this,
k+4/k−2 >0
And also i put −b/a value in the equation <0
Then used quadratic formula
64−4(k-2)(k+4) <0
And by trail and error,i got answer as 2.
Is there another way rather than intuition
We are given:
$\displaystyle (k-2)x^2+8x+(k+4)=0$
In order for the roots to be distinct and real, we require the discriminant to be positive:
$\displaystyle 8^2-4(k-2)(k+4)>0$
Solving this, we obtain:
$\displaystyle -6<k<4$
In order for the roots to be negative, we require:
$\displaystyle -8+\sqrt{8^2-4(k-2)(k+4)}<0$
Solving this, we obtain the intervals:
$\displaystyle [-6,-4)\,\cup\,(2,4]$
Putting the two together, we obtain:
$\displaystyle (-6,-4)\,\cup\,(2,4)$
And so we find (given $\displaystyle k\in\mathbb{Z}$):
$\displaystyle k\in\{-5,3\}$
Two values, as you found.
Yes, in my haste I didn't consider the sign of the denominator of the quadratic formula. So, we need to consider:
$\displaystyle \frac{-8+\sqrt{8^2-4(k-2)(k+4)}}{k-2}<0$
Which yields:
$\displaystyle (-4,2)\,\cup\,(2,4]$
And then:
$\displaystyle \frac{-8-\sqrt{8^2-4(k-2)(k+4)}}{k-2}<0$
Which yields:
$\displaystyle (2,4]$
Putting all the restrictions together, we get:
$\displaystyle (2,4)$
And so $\displaystyle k=3$ is the only solution satisfying the given conditions.