Determine all the values of the parameter m for which the operation: #4x^2-6mx+(2m+3)(m-3)# has TWO different real solutions. #x_1<x_2# #(4x_1-4x_2-1)(4x_1-4x_2+1)<0# ?

Question

#4x^2-6mx+(2m+3)(m-3)#

#x_1<x_2#

#(4x_1-4x_2-1)(4x_1-4x_2+1)<0#

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Answer 1 · 2017-05-12T11:58:09

#m < -13/2# or #-1/6 < m#

Solving for #x#

#4 x^2 - 6 m x + (2 m + 3) (m - 3) = 0#

we obtain

#{(x_1=1/2(m-3)),(x_2=1/2(2m+3)):}#

so

#(4 x_1 - 4 x_2 - 1) (4 x_1 + 4 x_2 + 1) = -(12m^2+80m+13) < 0#

so we will determine #m# such that

#12m^2+80m+13 > 0#

so

#m < -13/2# or #-1/6 < m#