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Question

If the equation ax2+2bx−3c=0 has non real roots and (3c/4)<(a+b); then C is always

A
<0
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B
>0
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C
0
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D
Zero
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Solution

The correct option is A <0
if the equation has non-real roots then,

(2b)24a(3c)<0

4b2+12ac<0

4b2<12ac , b2<3ac , a>b2/(3c) , a<b2/(3c)

12ac<0 (4b2 is a positive number)

either c<0 or a<0

given, 3c/4<(a+b)

3c/4<(b2/3c+b) (putting maximum value of a)

above equation implies,
c<0


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