Equation

Either -

x ² - 2 m x - 4 m ² - 4 = 0

Or ,

x ² - 4 x - 2 m ³ - 2 m = 0

Or , both are simultaneously zero , which in this case , cannot happen because it is stated that " f " has 3 distinct real roots .

See , the discriminant of the first equation , D = 20 m ² + 16 > 0 for all " m " .

Hence ,this equation will always have 2 distinct real roots .

So ,we require the second equation to have a repeated root .

For that , D = 0

Or , m ³ + m + 2 = 0 ,

Or , ( m ² - m + 2 ) ( m + 1 ) = 0 ......................... ( 1 )

which gives " m = - 1 " ,since the quadratic equation has only complex roots .

But , for " m = - 1 " ,

f ( x ) = ( x ² + 2 x - 8 ) ( x - 2 ) ²

= ( x - 2 ) ³ ( x + 4 )

which amazingly turns out to have only 2 real and different roots , not 3 , i . e , these equations have a common root .

Hence , no real values of " m " are there .

However , for complex values of " m " , you can always solve the quadratic equation in ( 1 ) .

yes i m also getting -1 as answer...
same method as ricky..