Consider f(x)=(x2-2mx-4(m2+1))(x2-4x-2m(m2+1))=0,m is not equal to 2.
If f(x) =0has three different real roots,then m=????
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2 Answers
1Either -
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 ) .
