+1124 Hi my good friends....please help me with this one..I have tried and tried...:
Determine the nature of the roots without solving the equation.
\(kx^2+(4k-1)x=2-4k\)
I did this....
\(kx^2+4kx-x+4k-2=0 \)
\(kx^2+x(4k-1)+2(2k-1)=0\)
Not sure..what I would not be suprised of is if I had made a mistake....
..
Thank you for all your help..always appreciated...
+1124 But then I did this: leave it as
\(kx^2+x(4k-1)+4k-2\)
\(d=b^2-4ac\)
\(=(4k-1)^2-4(k)(4k-2)\)
\(=16k^2-8k+1-16k^2+8k\)
\(=1\)
So the roots are Real, un-equal and rational....is this right perhaps?
+118608 Yes, that is correct. 
The symbol for the determinant is a triangle.
Backslash triangle. 
\(\triangle \)
+1124 Hello dear Melody.....
Thank you for confirming....yes I know its the triangle, just do not know where to get the symbol...I searched but could not find it...
