If the roots of the equation #bx^2 + cx + a = 0# be imaginary, then for all real values of x , the expression #3b^2x^2 + 6bcx + 2c^2# is?

Question

A: Greater than #-4ab#
B: Less than #-4ab#
C: Greater than #4ab#
D: Less than #4ab#
The answer is #(A)# for your reference

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Answer 1 · 2017-08-02T15:37:56

Given thar the roots of the equation #bx^2+cx+a=0# is imaginary,

So

#c^2-4ab<0#

#=>-c^2> -4ab......[1]#

Let

#3b^2x^2+6bcx+2c^2=y#

#=>3b^2x^2+6bcx+2c^2-y=0#

For real values of x we have

#(6bc)^2-4*(3b^2)*(2c^2-y)>=0#

#=>36b^2c^2-12*b^2(2c^2-y)>=0#

#=>3c^2-2c^2+y>=0#

#=>c^2+y>=0#

#=>y>=-c^2.....[2]#

comparing [1] and [2]

we get #y>(-4ab)#