1. ## Quadratic equations nature of roots

In question number 2,
I did like this,
k+4/k−2 >0
And also i put −b/a value in the equation <0
64−4(k-2)(k+4) <0
And by trail and error,i got answer as 2.
Is there another way rather than intuition

2. ## Re: Quadratic equations nature of roots

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.

3. ## Re: Quadratic equations nature of roots

for $\displaystyle k=-5$ the roots are positive

4. ## Re: Quadratic equations nature of roots

Originally Posted by Idea
for $\displaystyle k=-5$ the roots are positive
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.