x^2+kx+k>0
find all possible values of 'k'
2006-10-28 06:35:31 · 6 answers · asked by joejoe 1 in Science & Mathematics ➔ Mathematics
k> -x^2/(x+1)
2006-10-28 06:39:28 · answer #1 · answered by Anonymous · 4⤊ 0⤋
It depends on what type of curve it describe. If it is an elipsoid the second term k is 0.9/a where a=1 here and the constant k= 0.9a and the values can range from 0.1 to 0.9 any valu less than 1.
For hyperbloid any value > 1 and the same equations. For parabloid it is =1 and it is meaningless to do it for parabolas here sice a and k are equal. Your equation is another form of ax^2+bx+c where a,b and c are distinct. Bur in your equation b and c are equal which is very rare.
2006-10-28 14:35:21 · answer #2 · answered by Mathew C 5 · 0⤊ 0⤋
form o to infinity=all positive numbers and 0.
2006-10-28 14:15:21 · answer #3 · answered by Samir K 1 · 0⤊ 0⤋
K must be > or = -1
2006-10-28 14:11:13 · answer #4 · answered by DONALD H 1 · 0⤊ 0⤋
k can be infinate positive is x is positive and infinite negative if x is negative.
2006-10-28 13:39:32 · answer #5 · answered by Renegade Rhianna 3 · 0⤊ 0⤋
da eqn is true for any k>0 .....i.e. all da positive numbers ..
2006-10-28 13:39:15 · answer #6 · answered by sjv_ch 1 · 0⤊ 1⤋