I guess you mean (k-2)x2+8x+k+4=0.
When k=3 this becomes x2+8x+7=0=(x+7)(x+1), therefore k=3 is one solution.
(k-2)x2+8x+k+4=0 is the same as:
x2+(8/(k-2))x=-(k+4)/(k-2), complete the square:
x2+(8/(k-2))x+(4/(k-2))2=
(4/(k-2))2-(k+4)/(k-2)=16/(k-2)2-(k+4)/(k-2)=
(1/(k-2)2)(16-(k+4)(k-2))=
(1/(k-2)2)(16-k2-2k+8))=(24-2k-k2)/(k-2)2=(6+k)(4-k)/(k-2)2,
(x+4/(k-2))2=(6+k)(4-k)/(k-2)2,
x+4/(k-2)=±√[(6+k)(4-k)]/(k-2),
x=-4/(k-2)±√[(6+k)(4-k)]/(k-2).
The expression under the square root has to be a perfect square:
(6+k)(4-k)≥0 (for real numbers), so -6≤k≤4. Since k is an integer there are 11 possible values:
-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4. However k-2 must not be zero because x would be undefined, so there are only 10 values for k:
-6, -5, -4, -3, -2, -1, 0, 1, 3, 4. Not all of these will produce rational roots for x.
If only rational roots are required the following 6 values of k apply:
-6, -5, -4, -1, 3, 4.
| k |
k-2 |
√[(6+k)(4-k)] |
x root |
| -6 |
-8 |
0 |
1/2 |
| -5 |
-7 |
3 |
1/7, 1 |
| -4 |
-6 |
4 |
0, 4/3 |
| -1 |
-3 |
5 |
-1/3, 3 |
| 3 |
1 |
3 |
-1, -7 |
| 4 |
2 |
0 |
-2 |
This table shows the rational roots for the 6 possible values of k.