The roots of equation x3-6x+9=0 is
-6
-9
6
-3
Explanation for the correct option:
Find the roots of the given equation
Given equation,
x3-6x+9=0⇒x3+3x2-3x2-9x+3x+9=0⇒x2x+3-3xx+3+3x+3=0⇒x+3x2-3x+3=0
Now root of the quadratic equation x2-3x+3is
x=3±9-122=3±i32∵i=-1
Therefore, there is only one real root and that is -3
Hence, the correct option is D.
Find the non real roots of the cubic equation x3−6x−9=0