21let p be the only real root of the equation.
then we have p3-(a-4)(a+2)p - a = 0 -------[1ST]
now, (x-p) is a factor of x3 - (a-4)(a+2)x - a
we can write,
x3 - (a-4)(a+2)x - a
=(x-p)[x2 + px +p2 - (a-4)(a+2)]
Now, as p is the only root ,[x2 + px +p2 - (a-4)(a+2)] has discriminant less than zero
so, p2 - 4{p2 - (a-4)(a+2)} < 0
or, 4(a-4)(a+2) - 3p2 < 0
now if the above equation is true for min value of 3p2 (which is zero), then it is true for all real p
so 4(a-4)(a+2)<0
which gives, a lies between (-2) and 4
21
Shubhodip
·2010-11-29 05:55:07
smbdy check if i am correct....
341That's right. here's another take on the problem: x^3 -(a^2-2a-8)x -a = 0
has only one root implies that its derivative has at most one root.
i.e. g(x) = 3x^2 -(a^2-2a-8) has at most one real root. If a=4 or a=-2, then x=0 is the only real root. If (a^2-2a-8)>0 then we have two distinct roots. Hence (a^2-2a-8)<0
This is also sufficient as if (a^2-2a-8)<0, then the given cubic is monotonic and therefore has exactly one root.
Hence a \in [-2,4]
1its correct yr.... thanx......