Question

What are the integral values of k for which the equation `(k-2)x^2 + 8x + (k+4)=0)` has both roots real, distinct and negative ?

Answer 1

`-6 < k < 4`

Explanation:

For roots to be real, distinct and possibly negative, `Delta >0`

`Delta=b^2-4ac`
`Delta=8^2-4(k-2)(k+4)`
`Delta=64-4(k^2+2k-8)`
`Delta=64-4k^2-8k+32`
`Delta=96-4k^2-8k`

Since `Delta >0`,
`96-4k^2-8k >0`
`4k^2+8k-96 <0`
`(4k+24)(k-4) <0`
`4(k+6)(k-4) <0`

graph{y=4(x+6)(x-4) [-10, 10, -5, 5]}
From the graph above, we can see that the equation is true only when `-6 < k < 4`

Therefore,, only integers between `-6 < k < 4` can the roots be negative, distinct and real